Chapter 13 Limits and Derivatives (Additional Questions)

We are asked to evaluate the limit:

$\lim\limits_{x \to 2} (3x^2 - 5x + 4)$

The function $f(x) = 3x^2 - 5x + 4$ is a polynomial function.

Polynomial functions are continuous for all real numbers.

For a continuous function $f(x)$, the limit as $x$ approaches a value 'a' is equal to the function evaluated at 'a', i.e., $\lim\limits_{x \to a} f(x) = f(a)$.

Here, $a = 2$. We can find the limit by substituting $x=2$ into the expression:

$\lim\limits_{x \to 2} (3x^2 - 5x + 4) = 3(2)^2 - 5(2) + 4$

$= 3(4) - 10 + 4$

$= 12 - 10 + 4$

$= 2 + 4$

$= 6$

The value of the limit is 6.

Comparing this result with the given options, the correct option is (A) 6.

We are asked to evaluate the limit:

$\lim\limits_{x \to 1} \frac{x^2 - 1}{x - 1}$

If we substitute $x = 1$ directly into the expression, we get $\frac{1^2 - 1}{1 - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0}$, which is an indeterminate form.

To evaluate this limit, we can simplify the expression by factoring the numerator. The numerator $x^2 - 1$ is a difference of squares, which can be factored as $(x - 1)(x + 1)$.

So, the expression becomes:

$\frac{x^2 - 1}{x - 1} = \frac{(x - 1)(x + 1)}{x - 1}$

For $x \neq 1$, we can cancel the common factor $(x - 1)$ from the numerator and the denominator:

$\frac{\cancel{(x - 1)}(x + 1)}{\cancel{x - 1}} = x + 1 \quad \text{for } x \neq 1$

The limit of the original expression as $x$ approaches 1 is the same as the limit of the simplified expression $x + 1$ as $x$ approaches 1, because the two functions are identical for all values of $x$ except at $x = 1$.

Now, we evaluate the limit of $x + 1$ as $x$ approaches 1:

$\lim\limits_{x \to 1} (x + 1)$

Since $f(x) = x + 1$ is a polynomial function, it is continuous everywhere. Therefore, we can find the limit by direct substitution of $x = 1$:

$\lim\limits_{x \to 1} (x + 1) = 1 + 1 = 2$

The value of the limit is 2.

Comparing this result with the given options, the correct option is (C) 2.

We are asked to evaluate the limit:

$\lim\limits_{x \to 0} \frac{\sin x}{x}$

If we substitute $x = 0$ directly into the expression, we get $\frac{\sin(0)}{0} = \frac{0}{0}$, which is an indeterminate form.

This limit is a fundamental result in calculus, often referred to as a standard trigonometric limit.

Its value can be established using geometric arguments (like the Squeeze Theorem) or other methods like L'Hopital's Rule (though L'Hopital's Rule itself relies on the derivative of $\sin x$, which is often derived using this limit).

The value of this limit is a widely known standard result:

$\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$

The value of the limit is 1.

Comparing this result with the given options, the correct option is (B) 1.

The derivative of a function $f(x)$ with respect to $x$, denoted by $f'(x)$ or $\frac{df}{dx}$, is defined as the limit of the difference quotient as the increment in $x$ approaches zero.

Let $h$ be a small increment in $x$. The change in the function value is $f(x+h) - f(x)$.

The difference quotient is the ratio of the change in the function value to the change in $x$, which is $\frac{f(x+h) - f(x)}{h}$.

The derivative is the limit of this difference quotient as $h$ approaches 0:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Comparing this definition with the given options:

(A) $\lim\limits_{h \to 0} \frac{f(x+h) + f(x)}{h}$ - Incorrect, the numerator should be a difference.

(B) $\lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$ - Correct, this matches the definition of the derivative.

(C) $\lim\limits_{x \to 0} \frac{f(x+h) - f(x)}{h}$ - Incorrect, the limit should be as $h \to 0$, not $x \to 0$.

(D) $\lim\limits_{h \to 0} (f(x+h) - f(x))$ - Incorrect, the expression is missing the division by $h$ and this limit would typically be 0 for a continuous function.

Therefore, the correct definition of the derivative is given by option (B).

The correct option is (B) $\lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$.

We need to find the derivative of $f(x) = x^n$ using the first principle.

The derivative of $f(x)$ using the first principle is defined as:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Here, $f(x) = x^n$ and $f(x+h) = (x+h)^n$.

Substitute these into the definition:

$f'(x) = \lim\limits_{h \to 0} \frac{(x+h)^n - x^n}{h}$

We use the binomial expansion for $(x+h)^n$:
$(x+h)^n = x^n + \binom{n}{1}x^{n-1}h + \binom{n}{2}x^{n-2}h^2 + ... + \binom{n}{n}h^n$

$(x+h)^n = x^n + nx^{n-1}h + \frac{n(n-1)}{2!}x^{n-2}h^2 + ... + h^n$

Now, substitute this expansion into the numerator:

$(x+h)^n - x^n = (x^n + nx^{n-1}h + \frac{n(n-1)}{2!}x^{n-2}h^2 + ... + h^n) - x^n$

$= nx^{n-1}h + \frac{n(n-1)}{2!}x^{n-2}h^2 + ... + h^n$

Substitute this back into the limit expression:

$f'(x) = \lim\limits_{h \to 0} \frac{nx^{n-1}h + \frac{n(n-1)}{2!}x^{n-2}h^2 + ... + h^n}{h}$

Factor out $h$ from the numerator:

$f'(x) = \lim\limits_{h \to 0} \frac{h(nx^{n-1} + \frac{n(n-1)}{2!}x^{n-2}h + ... + h^{n-1})}{h}$

Since $h \to 0$ but $h \neq 0$, we can cancel $h$ from the numerator and denominator:

$f'(x) = \lim\limits_{h \to 0} (nx^{n-1} + \frac{n(n-1)}{2!}x^{n-2}h + ... + h^{n-1})$

Now, take the limit as $h \to 0$. All terms containing $h$ will become 0:

$f'(x) = nx^{n-1} + \frac{n(n-1)}{2!}x^{n-2}(0) + ... + (0)^{n-1}$

$f'(x) = nx^{n-1} + 0 + ... + 0$

$f'(x) = nx^{n-1}$

The derivative of $f(x) = x^n$ is $nx^{n-1}$.

Comparing this result with the given options, the correct option is (A) $nx^{n-1}$.

We are asked to evaluate the limit:

$\lim\limits_{x \to 0} \frac{\tan x}{x}$

If we substitute $x = 0$ directly into the expression, we get $\frac{\tan(0)}{0} = \frac{0}{0}$, which is an indeterminate form.

We can rewrite $\tan x$ as $\frac{\sin x}{\cos x}$.

So, the expression becomes:

$\frac{\tan x}{x} = \frac{\frac{\sin x}{\cos x}}{x} = \frac{\sin x}{x \cos x}$

We can separate this expression into a product of two parts:

$\frac{\sin x}{x \cos x} = \frac{\sin x}{x} \cdot \frac{1}{\cos x}$

Now, we can evaluate the limit using the property that the limit of a product is the product of the limits (provided the individual limits exist):

$\lim\limits_{x \to 0} \frac{\tan x}{x} = \lim\limits_{x \to 0} \left(\frac{\sin x}{x} \cdot \frac{1}{\cos x}\right)$

$= \left(\lim\limits_{x \to 0} \frac{\sin x}{x}\right) \cdot \left(\lim\limits_{x \to 0} \frac{1}{\cos x}\right)$

We know the standard limit:

$\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$

And the limit of $\frac{1}{\cos x}$ as $x \to 0$ is:

$\lim\limits_{x \to 0} \frac{1}{\cos x} = \frac{1}{\cos 0}$

Since $\cos 0 = 1$, we have:

$\lim\limits_{x \to 0} \frac{1}{\cos x} = \frac{1}{1} = 1$

Substitute these values back into the product of limits:

$\lim\limits_{x \to 0} \frac{\tan x}{x} = (1) \cdot (1) = 1$

The value of the limit is 1.

Comparing this result with the given options, the correct option is (B) 1.

Let's analyze the Assertion (A) and Reason (R).

Assertion (A): The limit of a function $f(x)$ as $x$ approaches a point 'a' exists if and only if the left-hand limit of $f(x)$ as $x$ approaches 'a' and the right-hand limit of $f(x)$ as $x$ approaches 'a' both exist and are equal to the same finite value.

This statement is a fundamental definition of the existence of a two-sided limit at a point. It is a true statement.

Reason (R): $\lim\limits_{x \to a} f(x)$ exists $\iff \lim\limits_{x \to a^-} f(x) = \lim\limits_{x \to a^+} f(x) = L$, where $L$ is a finite value.

This statement uses mathematical notation to express the same condition mentioned in Assertion (A). It formally states that the two-sided limit at 'a' exists if and only if the left-hand limit ($\lim\limits_{x \to a^-} f(x)$) equals the right-hand limit ($\lim\limits_{x \to a^+} f(x)$), and this common value is a finite number $L$. This statement is also a true statement and is the formal definition of the existence of a limit.

Comparing Assertion (A) and Reason (R), we see that Reason (R) provides the precise mathematical condition for the truth of Assertion (A). Reason (R) is the formal definition that explains why Assertion (A) is true.

Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A).

The correct option is (A) Both A and R are true and R is the correct explanation of A.

We are asked to evaluate the limit:

$\lim\limits_{x \to a} \frac{x^n - a^n}{x - a}$

If we substitute $x = a$ directly into the expression, we get $\frac{a^n - a^n}{a - a} = \frac{0}{0}$, which is an indeterminate form.

Method 1: Using Factorization

We can factor the numerator $x^n - a^n$ using the formula for the difference of powers:

$x^n - a^n = (x - a)(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \dots + xa^{n-2} + a^{n-1})$

Substitute this factorization into the limit expression:

$\lim\limits_{x \to a} \frac{(x - a)(x^{n-1} + x^{n-2}a + \dots + a^{n-1})}{x - a}$

For $x \neq a$, we can cancel the common factor $(x - a)$ from the numerator and the denominator:

$\lim\limits_{x \to a} (x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \dots + xa^{n-2} + a^{n-1})$

Now, evaluate the limit by substituting $x = a$ into the simplified expression. Since this is a polynomial in $x$, the limit can be found by direct substitution.

The expression is a sum of $n$ terms. Each term has the form $x^{n-k}a^{k-1}$ for $k=1, 2, \dots, n$.

As $x \to a$, each term $x^{n-k}a^{k-1}$ approaches $a^{n-k}a^{k-1} = a^{(n-k)+(k-1)} = a^{n-1}$.

There are $n$ such terms in the sum $(x^{n-1} + x^{n-2}a + \dots + a^{n-1})$.

Therefore, the limit of the sum is the sum of the limits:

$\lim\limits_{x \to a} (x^{n-1} + x^{n-2}a + \dots + a^{n-1}) = \lim\limits_{x \to a} x^{n-1} + \lim\limits_{x \to a} x^{n-2}a + \dots + \lim\limits_{x \to a} a^{n-1}$

$= a^{n-1} + a^{n-1} + a^{n-1} + \dots + a^{n-1}$ (n times)

$= n \cdot a^{n-1}$

Method 2: Using L'Hopital's Rule

Since substituting $x=a$ gives the indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule.

Let $f(x) = x^n - a^n$ and $g(x) = x - a$.

Find the derivatives of $f(x)$ and $g(x)$ with respect to $x$ (treating $a$ as a constant):

$f'(x) = \frac{d}{dx}(x^n - a^n) = nx^{n-1} - 0 = nx^{n-1}$

$g'(x) = \frac{d}{dx}(x - a) = 1 - 0 = 1$

Apply L'Hopital's Rule:

$\lim\limits_{x \to a} \frac{x^n - a^n}{x - a} = \lim\limits_{x \to a} \frac{f'(x)}{g'(x)} = \lim\limits_{x \to a} \frac{nx^{n-1}}{1}$

Now, substitute $x = a$ into the simplified expression:

$\lim\limits_{x \to a} nx^{n-1} = na^{n-1}$

Both methods yield the same result.

The value of the limit is $n a^{n-1}$. This is also a standard limit formula and is equivalent to the definition of the derivative of $f(x) = x^n$ at $x=a$.

Comparing this result with the given options, the correct option is (A) $n a^{n-1}$.

We need to find the derivative of each function listed in the left column and match it with the corresponding derivative in the right column.

(i) $f(x) = c$ (constant)

The derivative of a constant function is 0.

$\frac{d}{dx}(c) = 0$

This matches option (d) 0.

So, (i) $\to$ (d).

(ii) $f(x) = x$

Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ with $n=1$:

$\frac{d}{dx}(x) = 1 \cdot x^{1-1} = 1 \cdot x^0 = 1 \cdot 1 = 1$

This matches option (c) 1.

So, (ii) $\to$ (c).

(iii) $f(x) = \sin x$

The standard derivative of $\sin x$ is $\cos x$.

$\frac{d}{dx}(\sin x) = \cos x$

This matches option (a) $\cos x$.

So, (iii) $\to$ (a).

(iv) $f(x) = \cos x$

The standard derivative of $\cos x$ is $-\sin x$.

$\frac{d}{dx}(\cos x) = -\sin x$

This matches option (b) $-\sin x$.

So, (iv) $\to$ (b).

The correct matching is:

(i) $\to$ (d)

(ii) $\to$ (c)

(iii) $\to$ (a)

(iv) $\to$ (b)

Comparing this with the given options:

(A) (i)-(a), (ii)-(b), (iii)-(c), (iv)-(d) - Incorrect

(B) (i)-(d), (ii)-(c), (iii)-(a), (iv)-(b) - Correct

(C) (i)-(d), (ii)-(c), (iii)-(b), (iv)-(a) - Incorrect

(D) (i)-(a), (ii)-(b), (iii)-(d), (iv)-(c) - Incorrect

The correct option is (B).

We are asked to evaluate the limit:

$\lim\limits_{x \to 0} (1+x)^{1/x}$

As $x \to 0$, the base $(1+x) \to (1+0) = 1$.

As $x \to 0$, the exponent $\frac{1}{x}$ tends to $\infty$ (specifically, $\frac{1}{x} \to +\infty$ as $x \to 0^+$ and $\frac{1}{x} \to -\infty$ as $x \to 0^-$).

Thus, the limit is in the indeterminate form $1^\infty$.

This limit is a fundamental limit in calculus and serves as one of the definitions of the mathematical constant $e$, the base of the natural logarithm.

The value of this standard limit is:

$\lim\limits_{x \to 0} (1+x)^{1/x} = e$

The value of the limit is $e$.

Comparing this result with the given options, the correct option is (C) $e$.

We are asked to find the derivative of the function $f(x) = x^2 + 3x + 5$.

To find the derivative of this function, we can use the properties of differentiation, specifically the sum rule, constant multiple rule, power rule, and the rule for differentiating a constant.

The derivative of a sum of functions is the sum of their derivatives:

$\frac{d}{dx}(u(x) + v(x) + w(x)) = \frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx}$

The derivative of $f(x) = x^2 + 3x + 5$ is:

$f'(x) = \frac{d}{dx}(x^2 + 3x + 5)$

$f'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(3x) + \frac{d}{dx}(5)$

Now, we differentiate each term:

1. Derivative of $x^2$:

Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$, with $n=2$:

$\frac{d}{dx}(x^2) = 2x^{2-1} = 2x^1 = 2x$

2. Derivative of $3x$:

Using the constant multiple rule $\frac{d}{dx}(cu(x)) = c\frac{du}{dx}$ and the power rule for $x$ (where $n=1$):

$\frac{d}{dx}(3x) = 3 \frac{d}{dx}(x^1) = 3 \cdot 1x^{1-1} = 3 \cdot 1x^0 = 3 \cdot 1 = 3$

3. Derivative of 5:

Using the rule for the derivative of a constant $\frac{d}{dx}(c) = 0$:

$\frac{d}{dx}(5) = 0$

Now, combine the derivatives of the terms:

$f'(x) = 2x + 3 + 0$

$f'(x) = 2x + 3$

The derivative of $f(x) = x^2 + 3x + 5$ is $2x + 3$.

Comparing this result with the given options, the correct option is (A) $2x + 3$.

We are asked to evaluate the limit of a rational function as $x \to \infty$:

$\lim\limits_{x \to \infty} \frac{3x^2 + 2x + 1}{2x^2 - x + 5}$

As $x \to \infty$, both the numerator $(3x^2 + 2x + 1)$ and the denominator $(2x^2 - x + 5)$ tend towards infinity. This gives the indeterminate form $\frac{\infty}{\infty}$.

To evaluate the limit of a rational function as $x \to \infty$ (or $x \to -\infty$), we can divide every term in the numerator and the denominator by the highest power of $x$ present in the denominator. In this case, the highest power of $x$ in the denominator is $x^2$.

Divide the numerator and the denominator by $x^2$:

$\lim\limits_{x \to \infty} \frac{\frac{3x^2}{x^2} + \frac{2x}{x^2} + \frac{1}{x^2}}{\frac{2x^2}{x^2} - \frac{x}{x^2} + \frac{5}{x^2}}$

Simplify each term:

$\lim\limits_{x \to \infty} \frac{3 + \frac{2}{x} + \frac{1}{x^2}}{2 - \frac{1}{x} + \frac{5}{x^2}}$

Now, we evaluate the limit of each term as $x \to \infty$. We use the fact that for any positive integer $k$, $\lim\limits_{x \to \infty} \frac{c}{x^k} = 0$, where $c$ is a constant.

$\lim\limits_{x \to \infty} 3 = 3$

$\lim\limits_{x \to \infty} \frac{2}{x} = 0$

$\lim\limits_{x \to \infty} \frac{1}{x^2} = 0$

$\lim\limits_{x \to \infty} 2 = 2$

$\lim\limits_{x \to \infty} \frac{1}{x} = 0$

$\lim\limits_{x \to \infty} \frac{5}{x^2} = 0$

Substitute these limits back into the expression:

$\frac{\lim\limits_{x \to \infty} (3 + \frac{2}{x} + \frac{1}{x^2})}{\lim\limits_{x \to \infty} (2 - \frac{1}{x} + \frac{5}{x^2})} = \frac{3 + 0 + 0}{2 - 0 + 0}$

$= \frac{3}{2}$

The value of the limit is $\frac{3}{2}$.

Comparing this result with the given options, the correct option is (A) 3/2.

We need to find the derivative of the function $f(x) = \tan x$.

We can write $\tan x$ in terms of $\sin x$ and $\cos x$:

$f(x) = \tan x = \frac{\sin x}{\cos x}$

We will use the quotient rule for differentiation, which states that if $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}$.

Let $u(x) = \sin x$ and $v(x) = \cos x$.

Find the derivatives of $u(x)$ and $v(x)$:

$u'(x) = \frac{d}{dx}(\sin x) = \cos x$

$v'(x) = \frac{d}{dx}(\cos x) = -\sin x$

Apply the quotient rule:

$f'(x) = \frac{\cos x (\cos x) - \sin x (-\sin x)}{(\cos x)^2}$

$f'(x) = \frac{\cos^2 x + \sin^2 x}{\cos^2 x}$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ in the numerator:

$f'(x) = \frac{1}{\cos^2 x}$

Using the reciprocal identity $\frac{1}{\cos x} = \sec x$:

$f'(x) = \left(\frac{1}{\cos x}\right)^2 = \sec^2 x$

The derivative of $f(x) = \tan x$ is $\sec^2 x$.

Comparing this result with the given options, the correct option is (A) $\sec^2 x$.

The position of the particle at time $t$ is given by the function:

$s(t) = t^3 - 6t^2 + 9t + 5$

The instantaneous velocity of the particle is the derivative of the position function with respect to time $t$:

$v(t) = \frac{ds}{dt}$

We need to differentiate $s(t)$ with respect to $t$:

$v(t) = \frac{d}{dt}(t^3 - 6t^2 + 9t + 5)$

Using the sum and difference rules of differentiation, we can differentiate each term separately:

$v(t) = \frac{d}{dt}(t^3) - \frac{d}{dt}(6t^2) + \frac{d}{dt}(9t) + \frac{d}{dt}(5)$

Applying the power rule $\frac{d}{dt}(t^n) = nt^{n-1}$ and the rule for the derivative of a constant $\frac{d}{dt}(c) = 0$:

$\frac{d}{dt}(t^3) = 3t^{3-1} = 3t^2$

$\frac{d}{dt}(6t^2) = 6 \cdot \frac{d}{dt}(t^2) = 6 \cdot (2t^{2-1}) = 6 \cdot (2t) = 12t$

$\frac{d}{dt}(9t) = 9 \cdot \frac{d}{dt}(t^1) = 9 \cdot (1t^{1-1}) = 9 \cdot (1) = 9$

$\frac{d}{dt}(5) = 0$

Substitute these results back into the expression for $v(t)$:

$v(t) = 3t^2 - 12t + 9 + 0$

$v(t) = 3t^2 - 12t + 9$

The instantaneous velocity function is $v(t) = 3t^2 - 12t + 9$.

Comparing this result with the given options, the correct option is (A) $3t^2 - 12t + 9$.

From the previous question (Question 14), the instantaneous velocity function is:

$v(t) = 3t^2 - 12t + 9$

The instantaneous acceleration of the particle is the derivative of the velocity function with respect to time $t$:

$a(t) = \frac{dv}{dt}$

We need to differentiate $v(t)$ with respect to $t$:

$a(t) = \frac{d}{dt}(3t^2 - 12t + 9)$

Using the sum and difference rules of differentiation, we differentiate each term separately:

$a(t) = \frac{d}{dt}(3t^2) - \frac{d}{dt}(12t) + \frac{d}{dt}(9)$

Applying the power rule $\frac{d}{dt}(t^n) = nt^{n-1}$ and the rule for the derivative of a constant $\frac{d}{dt}(c) = 0$:

$\frac{d}{dt}(3t^2) = 3 \cdot \frac{d}{dt}(t^2) = 3 \cdot (2t^{2-1}) = 6t$

$\frac{d}{dt}(12t) = 12 \cdot \frac{d}{dt}(t^1) = 12 \cdot (1t^{1-1}) = 12 \cdot 1 = 12$

$\frac{d}{dt}(9) = 0$

Substitute these results back into the expression for $a(t)$:

$a(t) = 6t - 12 + 0$

$a(t) = 6t - 12$

The instantaneous acceleration function is $a(t) = 6t - 12$.

Comparing this result with the given options, the correct option is (A) $6t - 12$.

From Question 14, the instantaneous velocity function is given by:

$v(t) = 3t^2 - 12t + 9$

We are asked to find the time(s) at which the velocity is zero. This means we need to solve the equation $v(t) = 0$ for $t$:

$3t^2 - 12t + 9 = 0$

This is a quadratic equation. We can solve it by factoring. First, divide the entire equation by 3 to simplify it:

$\frac{3t^2}{3} - \frac{12t}{3} + \frac{9}{3} = \frac{0}{3}$

$t^2 - 4t + 3 = 0$

Now, factor the quadratic expression. We need two numbers that multiply to 3 and add up to -4. These numbers are -1 and -3.

So, we can write the equation as:

$(t - 1)(t - 3) = 0$

For the product of two factors to be zero, at least one of the factors must be zero. Therefore, we have two possible cases:

Case 1: $t - 1 = 0$

Case 2: $t - 3 = 0$

The velocity of the particle is zero at $t = 1$ and $t = 3$.

Comparing these values with the given options, the correct option is (A) $t=1, 3$.

We are asked to evaluate the limit:

$\lim\limits_{x \to 0} \frac{e^x - 1}{x}$

If we substitute $x = 0$ directly into the expression, we get $\frac{e^0 - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0}$, which is an indeterminate form.

This is a standard limit related to the definition of the derivative of $e^x$ at $x=0$.

Let $f(x) = e^x$. Then $f(0) = e^0 = 1$. The derivative of $f(x)$ at $x=0$ is given by the limit definition:

$f'(0) = \lim\limits_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim\limits_{h \to 0} \frac{e^{0+h} - e^0}{h} = \lim\limits_{h \to 0} \frac{e^h - 1}{h}$

Replacing $h$ with $x$, we get $\lim\limits_{x \to 0} \frac{e^x - 1}{x}$.

We also know that the derivative of $f(x) = e^x$ is $f'(x) = e^x$.

So, $f'(0) = e^0 = 1$.

Therefore, $\lim\limits_{x \to 0} \frac{e^x - 1}{x} = 1$.

Alternatively, we can use L'Hopital's Rule since the limit is in the $\frac{0}{0}$ indeterminate form.

Let $g(x) = e^x - 1$ and $h(x) = x$.

Find the derivatives of $g(x)$ and $h(x)$ with respect to $x$:

$g'(x) = \frac{d}{dx}(e^x - 1) = e^x - 0 = e^x$

$h'(x) = \frac{d}{dx}(x) = 1$

Apply L'Hopital's Rule:

$\lim\limits_{x \to 0} \frac{e^x - 1}{x} = \lim\limits_{x \to 0} \frac{g'(x)}{h'(x)} = \lim\limits_{x \to 0} \frac{e^x}{1}$

Now, substitute $x = 0$ into the simplified expression:

$\lim\limits_{x \to 0} e^x = e^0 = 1$

The value of the limit is 1.

Comparing this result with the given options, the correct option is (B) 1.

We are asked to find the derivative of the function $f(x) = \log_e x$.

The function $\log_e x$ is also commonly written as $\ln x$, which represents the natural logarithm of $x$.

There is a standard formula for the derivative of the natural logarithm function:

$\frac{d}{dx}(\log_e x) = \frac{d}{dx}(\ln x) = \frac{1}{x}$

This formula is typically derived using the definition of the derivative (first principle) or by using implicit differentiation on the inverse function $e^x$.

Applying this standard formula directly to the given function:

$f'(x) = \frac{d}{dx}(\log_e x) = \frac{1}{x}$

The derivative of $f(x) = \log_e x$ is $\frac{1}{x}$.

Comparing this result with the given options, the correct option is (B) $1/x$.

We are asked to evaluate the limit:

$\lim\limits_{x \to 0} \frac{a^x - 1}{x}$

If we substitute $x = 0$ directly into the expression, we get $\frac{a^0 - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0}$, which is an indeterminate form.

Since this is an indeterminate form $\frac{0}{0}$, we can use L'Hopital's Rule.

Let $f(x) = a^x - 1$ and $g(x) = x$.

Find the derivatives of $f(x)$ and $g(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(a^x - 1)$

The derivative of $a^x$ is $a^x \log_e a$. The derivative of a constant (1) is 0.

$f'(x) = a^x \log_e a - 0 = a^x \log_e a$

$g'(x) = \frac{d}{dx}(x) = 1$

Apply L'Hopital's Rule:

$\lim\limits_{x \to 0} \frac{a^x - 1}{x} = \lim\limits_{x \to 0} \frac{f'(x)}{g'(x)} = \lim\limits_{x \to 0} \frac{a^x \log_e a}{1}$

Now, evaluate the limit by substituting $x = 0$ into the simplified expression:

$\lim\limits_{x \to 0} (a^x \log_e a) = a^0 \cdot \log_e a$

Since $a^0 = 1$, we have:

$1 \cdot \log_e a = \log_e a$

The value of the limit is $\log_e a$.

Comparing this result with the given options, we see that option (C) is $\log a$. In the context of calculus and limits involving exponential functions with base $a$, $\log a$ typically refers to the natural logarithm, $\log_e a$ or $\ln a$. Assuming $\log a$ in option (C) means $\log_e a$, then option (C) is the correct answer.

The correct option is (C) $\log a$ (interpreting $\log a$ as $\log_e a$).

We need to identify the formula that represents the product rule for differentiation from the given options.

Let's examine each option:

(A) $\frac{d}{dx}(u+v) = \frac{du}{dx} + \frac{dv}{dx}$

This is the Sum Rule for differentiation. It states that the derivative of a sum of two functions is the sum of their derivatives.

(B) $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$

This is the Product Rule for differentiation. It states that the derivative of the product of two functions $u$ and $v$ is the first function times the derivative of the second, plus the second function times the derivative of the first.

(C) $\frac{d}{dx}(\frac{u}{v}) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

This is the Quotient Rule for differentiation. It states that the derivative of a quotient of two functions is the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator.

(D) $\frac{d}{dx}(u^n) = n u^{n-1} \frac{du}{dx}$

This is the Chain Rule applied to the power rule. It is used to differentiate a function raised to a power, where the base is itself a function of $x$. If $u=f(x)$ and $y=u^n$, then $\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = nu^{n-1}\frac{du}{dx}$.

Based on the definitions, the formula representing the product rule is option (B).

The correct option is (B) $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$.

We are asked to find the derivative of the function $f(x) = (x^2 + 1)(x^3 - 2)$.

This function is a product of two functions. We can use the product rule for differentiation.

Let $u(x) = x^2 + 1$ and $v(x) = x^3 - 2$.

The product rule states that if $f(x) = u(x)v(x)$, then $f'(x) = u(x)v'(x) + v(x)u'(x)$, where $u'(x) = \frac{du}{dx}$ and $v'(x) = \frac{dv}{dx}$.

First, find the derivative of $u(x)$ with respect to $x$:

$u'(x) = \frac{d}{dx}(x^2 + 1)$

$u'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(1)$

$u'(x) = 2x + 0$

$u'(x) = 2x$

Next, find the derivative of $v(x)$ with respect to $x$:

$v'(x) = \frac{d}{dx}(x^3 - 2)$

$v'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(2)$

$v'(x) = 3x^2 - 0$

$v'(x) = 3x^2$

Now, apply the product rule formula $f'(x) = u(x)v'(x) + v(x)u'(x)$:

$f'(x) = (x^2 + 1)(3x^2) + (x^3 - 2)(2x)$

This expression matches option (A).

We can also simplify this expression further:

$f'(x) = 3x^2(x^2) + 3x^2(1) + 2x(x^3) + 2x(-2)$

$f'(x) = 3x^4 + 3x^2 + 2x^4 - 4x$

$f'(x) = (3x^4 + 2x^4) + 3x^2 - 4x$

$f'(x) = 5x^4 + 3x^2 - 4x$

While the simplified form is $5x^4 + 3x^2 - 4x$, option (A) is the correct application of the product rule before simplification, and it is listed as one of the options.

Comparing the results with the given options, option (A) is a correct representation of the derivative.

The correct option is (A) $2x(x^3-2) + 3x^2(x^2+1)$.

We are asked to evaluate the limit:

$\lim\limits_{x \to 3} \frac{x^2 - 9}{x - 3}$

If we substitute $x = 3$ directly into the expression, we get $\frac{3^2 - 9}{3 - 3} = \frac{9 - 9}{0} = \frac{0}{0}$, which is an indeterminate form.

To evaluate this limit, we can simplify the expression by factoring the numerator. The numerator $x^2 - 9$ is a difference of squares, which can be factored as $(x - 3)(x + 3)$.

So, the expression becomes:

$\frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3}$

For $x \neq 3$, we can cancel the common factor $(x - 3)$ from the numerator and the denominator:

$\frac{\cancel{(x - 3)}(x + 3)}{\cancel{x - 3}} = x + 3 \quad \text{for } x \neq 3$

The limit of the original expression as $x$ approaches 3 is the same as the limit of the simplified expression $x + 3$ as $x$ approaches 3, because the two functions are identical for all values of $x$ except at $x = 3$.

Now, we evaluate the limit of $x + 3$ as $x$ approaches 3:

$\lim\limits_{x \to 3} (x + 3)$

Since $f(x) = x + 3$ is a polynomial function, it is continuous everywhere. Therefore, we can find the limit by direct substitution of $x = 3$:

$\lim\limits_{x \to 3} (x + 3) = 3 + 3 = 6$

The value of the limit is 6.

Comparing this result with the given options, the correct option is (C) 6.

We are asked to find the derivative of the function $f(x) = \frac{x+1}{x-1}$.

This function is a quotient of two functions. We can use the quotient rule for differentiation.

Let $u(x) = x+1$ (the numerator) and $v(x) = x-1$ (the denominator).

The quotient rule states that if $f(x) = \frac{u(x)}{v(x)}$, then the derivative $f'(x)$ is given by:

$f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}$

First, find the derivative of $u(x)$ with respect to $x$:

$u'(x) = \frac{d}{dx}(x+1)$

$u'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(1)$

$u'(x) = 1 + 0$

$u'(x) = 1$

Next, find the derivative of $v(x)$ with respect to $x$:

$v'(x) = \frac{d}{dx}(x-1)$

$v'(x) = \frac{d}{dx}(x) - \frac{d}{dx}(1)$

$v'(x) = 1 - 0$

$v'(x) = 1$

Now, apply the quotient rule formula $f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}$:

$f'(x) = \frac{(x-1)(1) - (x+1)(1)}{(x-1)^2}$

Simplify the numerator:

$(x-1)(1) - (x+1)(1) = (x-1) - (x+1) = x - 1 - x - 1 = -2$

Substitute the simplified numerator back into the expression for $f'(x)$:

$f'(x) = \frac{-2}{(x-1)^2}$

The derivative of $f(x) = \frac{x+1}{x-1}$ is $\frac{-2}{(x-1)^2}$.

Comparing this result with the given options, the correct option is (A) $\frac{(x-1)(1) - (x+1)(1)}{(x-1)^2} = \frac{x-1-x-1}{(x-1)^2} = \frac{-2}{(x-1)^2}$.

We are asked to evaluate the limit:

$\lim\limits_{x \to 0} \frac{1 - \cos x}{x^2}$

If we substitute $x = 0$ directly into the expression, we get $\frac{1 - \cos 0}{0^2} = \frac{1 - 1}{0} = \frac{0}{0}$, which is an indeterminate form.

Method 1: Using Trigonometric Identity and Standard Limit

We use the half-angle identity: $1 - \cos x = 2 \sin^2 \left(\frac{x}{2}\right)$.

Substitute this into the limit expression:

$\lim\limits_{x \to 0} \frac{2 \sin^2 \left(\frac{x}{2}\right)}{x^2}$

Rewrite the denominator to match the argument of the sine function squared:

$x^2 = \left(2 \cdot \frac{x}{2}\right)^2 = 4 \left(\frac{x}{2}\right)^2$

Substitute this back into the limit:

$\lim\limits_{x \to 0} \frac{2 \sin^2 \left(\frac{x}{2}\right)}{4 \left(\frac{x}{2}\right)^2}$

$= \lim\limits_{x \to 0} \frac{2}{4} \cdot \frac{\sin^2 \left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^2}$

$= \frac{1}{2} \lim\limits_{x \to 0} \left(\frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}}\right)^2$

Let $\theta = \frac{x}{2}$. As $x \to 0$, $\theta \to \frac{0}{2} = 0$. The expression becomes:

$= \frac{1}{2} \lim\limits_{\theta \to 0} \left(\frac{\sin \theta}{\theta}\right)^2$

Using the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

$= \frac{1}{2} (1)^2$

$= \frac{1}{2} \cdot 1 = \frac{1}{2}$

Method 2: Using L'Hopital's Rule

Since substituting $x=0$ gives the indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule.

Let $f(x) = 1 - \cos x$ and $g(x) = x^2$.

Find the derivatives of $f(x)$ and $g(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(1 - \cos x) = 0 - (-\sin x) = \sin x$

$g'(x) = \frac{d}{dx}(x^2) = 2x$

Apply L'Hopital's Rule:

$\lim\limits_{x \to 0} \frac{1 - \cos x}{x^2} = \lim\limits_{x \to 0} \frac{f'(x)}{g'(x)} = \lim\limits_{x \to 0} \frac{\sin x}{2x}$

If we substitute $x=0$ again, we still get $\frac{\sin 0}{2(0)} = \frac{0}{0}$, which is an indeterminate form. So, we apply L'Hopital's Rule again.

Find the derivatives of the new numerator and denominator:

$f''(x) = \frac{d}{dx}(\sin x) = \cos x$

$g''(x) = \frac{d}{dx}(2x) = 2$

Apply L'Hopital's Rule again:

$\lim\limits_{x \to 0} \frac{\sin x}{2x} = \lim\limits_{x \to 0} \frac{f''(x)}{g''(x)} = \lim\limits_{x \to 0} \frac{\cos x}{2}$

Now, substitute $x = 0$ into the simplified expression:

$\lim\limits_{x \to 0} \frac{\cos x}{2} = \frac{\cos 0}{2} = \frac{1}{2}$

Both methods yield the same result.

The value of the limit is $\frac{1}{2}$.

Comparing this result with the given options, the correct option is (B) 1/2.

We are asked to provide the derivative of a constant function.

A constant function is a function of the form $f(x) = c$, where $c$ is a real number. This means the value of the function does not change as $x$ changes.

The derivative of a function represents the instantaneous rate of change of the function with respect to its independent variable. If the function value does not change, its rate of change is zero.

Using the power rule for differentiation, the derivative of $f(x) = c$ can be thought of as $f(x) = c \cdot x^0$. Although this isn't the standard way to derive the derivative of a constant, the formal definition of the derivative using the first principle gives:

Let $f(x) = c$. Then $f(x+h) = c$.

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

$f'(x) = \lim\limits_{h \to 0} \frac{c - c}{h}$

$f'(x) = \lim\limits_{h \to 0} \frac{0}{h}$

For $h \neq 0$, $\frac{0}{h} = 0$.

$f'(x) = \lim\limits_{h \to 0} 0 = 0$

Thus, the derivative of any constant function is 0.

Comparing this result with the given options, the correct option is (B) 0.

Let's analyze the Assertion (A) and Reason (R).

Assertion (A): The derivative of $f(x) = \cos x$ is $f'(x) = -\sin x$.

This is a standard differentiation formula and is true.

Reason (R): The reason provides the derivation of the derivative of $f(x) = \cos x$ using the first principle definition of the derivative.

The derivative of $f(x)$ is given by $f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$.

For $f(x) = \cos x$, we have $f(x+h) = \cos(x+h)$.

So, $f'(x) = \lim\limits_{h \to 0} \frac{\cos(x+h) - \cos x}{h}$.

Using the trigonometric identity $\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$, with $A = x+h$ and $B = x$, we get:

$\cos(x+h) - \cos x = -2 \sin\left(\frac{(x+h)+x}{2}\right)\sin\left(\frac{(x+h)-x}{2}\right)$

$= -2 \sin\left(\frac{2x+h}{2}\right)\sin\left(\frac{h}{2}\right)$

$= -2 \sin\left(x + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right)$

Substitute this back into the limit:

$f'(x) = \lim\limits_{h \to 0} \frac{-2 \sin(x+h/2)\sin(h/2)}{h}$

We can rewrite this as:

$f'(x) = \lim\limits_{h \to 0} \left(-2 \sin\left(x + \frac{h}{2}\right) \cdot \frac{\sin(h/2)}{h}\right)$

$f'(x) = \lim\limits_{h \to 0} \left(-2 \sin\left(x + \frac{h}{2}\right) \cdot \frac{\sin(h/2)}{2 \cdot (h/2)}\right)$

$f'(x) = \lim\limits_{h \to 0} \left(- \sin\left(x + \frac{h}{2}\right) \cdot \frac{\sin(h/2)}{h/2}\right)$

Using the property that the limit of a product is the product of the limits (if they exist):

$f'(x) = \left(\lim\limits_{h \to 0} - \sin\left(x + \frac{h}{2}\right)\right) \cdot \left(\lim\limits_{h \to 0} \frac{\sin(h/2)}{h/2}\right)$

Evaluate the individual limits:

As $h \to 0$, $h/2 \to 0$. Since $\sin$ is a continuous function, $\lim\limits_{h \to 0} \sin\left(x + \frac{h}{2}\right) = \sin(x + 0) = \sin x$.

Using the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, where $\theta = h/2$, we have $\lim\limits_{h \to 0} \frac{\sin(h/2)}{h/2} = 1$.

Substitute these values back:

$f'(x) = (-\sin x) \cdot (1)$

$f'(x) = -\sin x$

The derivation shown in Reason (R) is correct and demonstrates how the derivative of $\cos x$ is $-\sin x$. Therefore, Reason (R) is a correct explanation of Assertion (A).

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

The correct option is (A) Both A and R are true and R is the correct explanation of A.

We are asked to evaluate the limit:

$\lim\limits_{x \to 0} \frac{\sin 3x}{\sin 5x}$

If we substitute $x = 0$ directly into the expression, we get $\frac{\sin (3 \cdot 0)}{\sin (5 \cdot 0)} = \frac{\sin 0}{\sin 0} = \frac{0}{0}$, which is an indeterminate form.

To evaluate this limit, we can use the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

We can rewrite the given expression by multiplying and dividing the numerator and the denominator by appropriate terms to match the form of the standard limit.

Multiply and divide the numerator by $3x$, and the denominator by $5x$:

$\frac{\sin 3x}{\sin 5x} = \frac{\frac{\sin 3x}{3x} \cdot 3x}{\frac{\sin 5x}{5x} \cdot 5x}$

$= \frac{\frac{\sin 3x}{3x}}{\frac{\sin 5x}{5x}} \cdot \frac{3x}{5x}$

$= \frac{\frac{\sin 3x}{3x}}{\frac{\sin 5x}{5x}} \cdot \frac{3}{5}$

Now, take the limit as $x \to 0$. We can apply the limit properties for quotients and products:

$\lim\limits_{x \to 0} \frac{\sin 3x}{\sin 5x} = \lim\limits_{x \to 0} \left(\frac{\frac{\sin 3x}{3x}}{\frac{\sin 5x}{5x}} \cdot \frac{3}{5}\right)$

$= \frac{\lim\limits_{x \to 0} \frac{\sin 3x}{3x}}{\lim\limits_{x \to 0} \frac{\sin 5x}{5x}} \cdot \lim\limits_{x \to 0} \frac{3}{5}$

For the terms involving sine, let $\theta_1 = 3x$ and $\theta_2 = 5x$. As $x \to 0$, both $\theta_1 \to 0$ and $\theta_2 \to 0$.

Using the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

$\lim\limits_{x \to 0} \frac{\sin 3x}{3x} = \lim\limits_{\theta_1 \to 0} \frac{\sin \theta_1}{\theta_1} = 1$

$\lim\limits_{x \to 0} \frac{\sin 5x}{5x} = \lim\limits_{\theta_2 \to 0} \frac{\sin \theta_2}{\theta_2} = 1$

$\lim\limits_{x \to 0} \frac{3}{5} = \frac{3}{5}$

Substitute these values back into the limit expression:

$\lim\limits_{x \to 0} \frac{\sin 3x}{\sin 5x} = \frac{1}{1} \cdot \frac{3}{5} = 1 \cdot \frac{3}{5} = \frac{3}{5}$

The value of the limit is $\frac{3}{5}$.

Comparing this result with the given options, the correct option is (A) 3/5.

We are asked to find the derivative of the function $f(x) = x \sin x$.

This function is a product of two functions: $u(x) = x$ and $v(x) = \sin x$.

We will use the product rule for differentiation.

The product rule states that if $f(x) = u(x)v(x)$, then the derivative $f'(x)$ is given by:

$f'(x) = u(x)v'(x) + v(x)u'(x)$

Let $u(x) = x$. Find its derivative with respect to $x$:

$u'(x) = \frac{d}{dx}(x) = 1$

Let $v(x) = \sin x$. Find its derivative with respect to $x$:

$v'(x) = \frac{d}{dx}(\sin x) = \cos x$

Now, apply the product rule formula $f'(x) = u(x)v'(x) + v(x)u'(x)$:

$f'(x) = (x)(\cos x) + (\sin x)(1)$

$f'(x) = x \cos x + \sin x$

The derivative of $f(x) = x \sin x$ is $x \cos x + \sin x$, which can also be written as $\sin x + x \cos x$.

Comparing this result with the given options, the correct option is (A) $\sin x + x \cos x$.

We are asked to evaluate the limit:

$\lim\limits_{x \to -1} \frac{x^3 + 1}{x + 1}$

If we substitute $x = -1$ directly into the expression, we get $\frac{(-1)^3 + 1}{-1 + 1} = \frac{-1 + 1}{0} = \frac{0}{0}$, which is an indeterminate form.

Method 1: Using Factorization

We can factor the numerator $x^3 + 1$ using the sum of cubes formula: $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$. Here, $a=x$ and $b=1$.

$x^3 + 1 = x^3 + 1^3 = (x+1)(x^2 - x(1) + 1^2) = (x+1)(x^2 - x + 1)$

Substitute this factorization into the limit expression:

$\lim\limits_{x \to -1} \frac{(x+1)(x^2 - x + 1)}{x + 1}$

For $x \neq -1$, we can cancel the common factor $(x+1)$ from the numerator and the denominator:

$\lim\limits_{x \to -1} \frac{\cancel{(x+1)}(x^2 - x + 1)}{\cancel{x + 1}} = \lim\limits_{x \to -1} (x^2 - x + 1) \quad \text{for } x \neq -1$

The limit of the original expression as $x$ approaches -1 is the same as the limit of the simplified expression $x^2 - x + 1$ as $x$ approaches -1.

Now, we evaluate the limit of $x^2 - x + 1$ as $x$ approaches -1. Since this is a polynomial function, it is continuous everywhere, so we can find the limit by direct substitution of $x = -1$:

$\lim\limits_{x \to -1} (x^2 - x + 1) = (-1)^2 - (-1) + 1$

$= 1 + 1 + 1$

$= 3$

Method 2: Using L'Hopital's Rule

Since substituting $x=-1$ gives the indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule.

Let $f(x) = x^3 + 1$ and $g(x) = x + 1$.

Find the derivatives of $f(x)$ and $g(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(x^3 + 1) = 3x^{3-1} + 0 = 3x^2$

$g'(x) = \frac{d}{dx}(x + 1) = 1 + 0 = 1$

Apply L'Hopital's Rule:

$\lim\limits_{x \to -1} \frac{x^3 + 1}{x + 1} = \lim\limits_{x \to -1} \frac{f'(x)}{g'(x)} = \lim\limits_{x \to -1} \frac{3x^2}{1}$

$= \lim\limits_{x \to -1} 3x^2$

Now, substitute $x = -1$ into the simplified expression:

$3(-1)^2 = 3(1) = 3$

Both methods yield the same result.

The value of the limit is 3.

Comparing this result with the given options, the correct option is (C) 3.

We are asked to find the derivative of the function $f(x) = e^x$.

The function $f(x) = e^x$ is the exponential function with base $e$.

The derivative of the exponential function $e^x$ with respect to $x$ is a fundamental result in calculus. It is equal to the function itself.

$\frac{d}{dx}(e^x) = e^x$

This can be shown using the definition of the derivative (first principle):

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

$f'(x) = \lim\limits_{h \to 0} \frac{e^{x+h} - e^x}{h}$

$f'(x) = \lim\limits_{h \to 0} \frac{e^x e^h - e^x}{h}$

$f'(x) = \lim\limits_{h \to 0} e^x \left(\frac{e^h - 1}{h}\right)$

$f'(x) = e^x \lim\limits_{h \to 0} \left(\frac{e^h - 1}{h}\right)$

Using the standard limit $\lim\limits_{h \to 0} \frac{e^h - 1}{h} = 1$, we get:

$f'(x) = e^x \cdot 1$

$f'(x) = e^x$

The derivative of $f(x) = e^x$ is $e^x$.

Comparing this result with the given options, option (B) is $e^x$. Option (C) is $e^x \log e$. If $\log e$ is interpreted as $\log_e e$ (natural logarithm), then $\log_e e = 1$, and option (C) becomes $e^x \cdot 1 = e^x$, which is the same as option (B).

The most direct and standard answer is $e^x$.

The correct option is (B) $e^x$.

We are asked to identify the standard limits from the given options.

Let's examine each option:

(A) $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$

This is a fundamental trigonometric limit. Its value is indeed 1.

This is a standard limit.

(B) $\lim\limits_{x \to 0} \frac{1 - \cos x}{x} = 0$

If we substitute $x=0$, we get $\frac{1 - \cos 0}{0} = \frac{1-1}{0} = \frac{0}{0}$, which is an indeterminate form.

Using L'Hopital's Rule:

$\lim\limits_{x \to 0} \frac{\frac{d}{dx}(1 - \cos x)}{\frac{d}{dx}(x)} = \lim\limits_{x \to 0} \frac{\sin x}{1}$

Substitute $x=0$: $\frac{\sin 0}{1} = \frac{0}{1} = 0$.

This is a standard limit, and its value is 0.

(C) $\lim\limits_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$

This is a standard algebraic limit, which is equivalent to the derivative of $x^n$ evaluated at $x=a$. Its value is indeed $n a^{n-1}$.

This is a standard limit.

(D) $\lim\limits_{x \to 0} \frac{\tan x}{x} = 1$

We can write $\frac{\tan x}{x} = \frac{\sin x}{x \cos x} = \frac{\sin x}{x} \cdot \frac{1}{\cos x}$.

$\lim\limits_{x \to 0} \frac{\tan x}{x} = \lim\limits_{x \to 0} \frac{\sin x}{x} \cdot \lim\limits_{x \to 0} \frac{1}{\cos x} = 1 \cdot \frac{1}{\cos 0} = 1 \cdot \frac{1}{1} = 1$.

This is a standard trigonometric limit, and its value is 1.

All four options provided are indeed standard limits.

The correct options are (A), (B), (C), and (D).

We are asked to evaluate the limit:

$\lim\limits_{x \to 0} \frac{\sin ax}{\sin bx}$

where $b \neq 0$.

If we substitute $x = 0$ directly into the expression, we get $\frac{\sin (a \cdot 0)}{\sin (b \cdot 0)} = \frac{\sin 0}{\sin 0} = \frac{0}{0}$, which is an indeterminate form.

To evaluate this limit, we can use the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

We can rewrite the given expression by multiplying and dividing the numerator and the denominator by appropriate terms to match the form of the standard limit.

Multiply and divide the numerator by $ax$ and the denominator by $bx$:

$\frac{\sin ax}{\sin bx} = \frac{\frac{\sin ax}{ax} \cdot ax}{\frac{\sin bx}{bx} \cdot bx}$

$= \frac{\frac{\sin ax}{ax}}{\frac{\sin bx}{bx}} \cdot \frac{ax}{bx}$

$= \frac{\frac{\sin ax}{ax}}{\frac{\sin bx}{bx}} \cdot \frac{a}{b} \quad$ (assuming $x \neq 0$)

Now, take the limit as $x \to 0$. We can apply the limit properties for quotients and products:

$\lim\limits_{x \to 0} \frac{\sin ax}{\sin bx} = \lim\limits_{x \to 0} \left(\frac{\frac{\sin ax}{ax}}{\frac{\sin bx}{bx}} \cdot \frac{a}{b}\right)$

$= \frac{\lim\limits_{x \to 0} \frac{\sin ax}{ax}}{\lim\limits_{x \to 0} \frac{\sin bx}{bx}} \cdot \lim\limits_{x \to 0} \frac{a}{b}$

For the terms involving sine, let $\theta_1 = ax$ and $\theta_2 = bx$. As $x \to 0$, both $\theta_1 \to 0$ and $\theta_2 \to 0$.

Using the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

$\lim\limits_{x \to 0} \frac{\sin ax}{ax} = \lim\limits_{\theta_1 \to 0} \frac{\sin \theta_1}{\theta_1} = 1 \quad$ (if $a \neq 0$)

$\lim\limits_{x \to 0} \frac{\sin bx}{bx} = \lim\limits_{\theta_2 \to 0} \frac{\sin \theta_2}{\theta_2} = 1 \quad$ (since $b \neq 0$)

$\lim\limits_{x \to 0} \frac{a}{b} = \frac{a}{b} \quad$ (since $a/b$ is a constant with respect to $x$)

Substitute these values back into the limit expression:

$\lim\limits_{x \to 0} \frac{\sin ax}{\sin bx} = \frac{1}{1} \cdot \frac{a}{b} = 1 \cdot \frac{a}{b} = \frac{a}{b}$

Note: If $a=0$, the limit becomes $\lim\limits_{x \to 0} \frac{\sin 0}{\sin bx} = \lim\limits_{x \to 0} \frac{0}{\sin bx}$. For $x$ close to 0 but $x \neq 0$, $\sin bx \neq 0$ (since $b \neq 0$), so the expression is 0. The limit is 0. Our derived formula $a/b$ gives $0/b = 0$, so the formula holds even when $a=0$.

The value of the limit is $\frac{a}{b}$.

Comparing this result with the given options, the correct option is (A) $a/b$.

We are asked to find the derivative of the function $f(x) = \text{cosec } x$.

The function $\text{cosec } x$ is the reciprocal of $\sin x$.

$\text{cosec } x = \frac{1}{\sin x}$

We can use the quotient rule to find the derivative. Let $u(x) = 1$ and $v(x) = \sin x$.

The quotient rule states that if $f(x) = \frac{u(x)}{v(x)}$, then the derivative $f'(x)$ is given by:

$f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}$

First, find the derivative of $u(x) = 1$ with respect to $x$:

$u'(x) = \frac{d}{dx}(1) = 0$

Next, find the derivative of $v(x) = \sin x$ with respect to $x$:

$v'(x) = \frac{d}{dx}(\sin x) = \cos x$

Now, apply the quotient rule formula $f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}$:

$f'(x) = \frac{(\sin x)(0) - (1)(\cos x)}{(\sin x)^2}$

$f'(x) = \frac{0 - \cos x}{\sin^2 x}$

$f'(x) = \frac{-\cos x}{\sin^2 x}$

We can rewrite this expression using trigonometric identities: $\sin^2 x = \sin x \cdot \sin x$, $\cot x = \frac{\cos x}{\sin x}$, and $\text{cosec } x = \frac{1}{\sin x}$.

$f'(x) = -\frac{\cos x}{\sin x} \cdot \frac{1}{\sin x}$

$f'(x) = -(\cot x) \cdot (\text{cosec } x)$

$f'(x) = -\text{cosec } x \cot x$

The derivative of $f(x) = \text{cosec } x$ is $-\text{cosec } x \cot x$. This is a standard differentiation formula.

Comparing this result with the given options, the correct option is (B) $-\text{cosec } x \cot x$.

We are asked to evaluate the limit:

$\lim\limits_{x \to 0} \frac{\sqrt{1+x} - 1}{x}$

If we substitute $x = 0$ directly into the expression, we get $\frac{\sqrt{1+0} - 1}{0} = \frac{\sqrt{1} - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0}$, which is an indeterminate form.

Method 1: Using Rationalization

To evaluate this limit, we can rationalize the numerator by multiplying the numerator and the denominator by the conjugate of the numerator, which is $\sqrt{1+x} + 1$.

$\lim\limits_{x \to 0} \frac{\sqrt{1+x} - 1}{x} \cdot \frac{\sqrt{1+x} + 1}{\sqrt{1+x} + 1}$

$= \lim\limits_{x \to 0} \frac{(\sqrt{1+x})^2 - (1)^2}{x(\sqrt{1+x} + 1)}$

$= \lim\limits_{x \to 0} \frac{(1+x) - 1}{x(\sqrt{1+x} + 1)}$

$= \lim\limits_{x \to 0} \frac{x}{x(\sqrt{1+x} + 1)}$

For $x \neq 0$, we can cancel the common factor $x$ from the numerator and the denominator:

$= \lim\limits_{x \to 0} \frac{\cancel{x}}{\cancel{x}(\sqrt{1+x} + 1)}$

$= \lim\limits_{x \to 0} \frac{1}{\sqrt{1+x} + 1}$

Now, evaluate the limit by substituting $x = 0$ into the simplified expression. The denominator is continuous at $x=0$, and the denominator is not zero at $x=0$.

$= \frac{1}{\sqrt{1+0} + 1}$

$= \frac{1}{\sqrt{1} + 1}$

$= \frac{1}{1 + 1} = \frac{1}{2}$

Method 2: Using L'Hopital's Rule

Since substituting $x=0$ gives the indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule.

Let $f(x) = \sqrt{1+x} - 1 = (1+x)^{1/2} - 1$ and $g(x) = x$.

Find the derivatives of $f(x)$ and $g(x)$ with respect to $x$. Use the chain rule for $f(x)$:

$f'(x) = \frac{d}{dx}((1+x)^{1/2} - 1) = \frac{1}{2}(1+x)^{1/2 - 1} \cdot \frac{d}{dx}(1+x) - 0$

$f'(x) = \frac{1}{2}(1+x)^{-1/2} \cdot (1 + 0)$

$f'(x) = \frac{1}{2}(1+x)^{-1/2} = \frac{1}{2\sqrt{1+x}}$

$g'(x) = \frac{d}{dx}(x) = 1$

Apply L'Hopital's Rule:

$\lim\limits_{x \to 0} \frac{\sqrt{1+x} - 1}{x} = \lim\limits_{x \to 0} \frac{f'(x)}{g'(x)} = \lim\limits_{x \to 0} \frac{\frac{1}{2\sqrt{1+x}}}{1}$

$= \lim\limits_{x \to 0} \frac{1}{2\sqrt{1+x}}$

Now, substitute $x = 0$ into the simplified expression:

$= \frac{1}{2\sqrt{1+0}} = \frac{1}{2\sqrt{1}} = \frac{1}{2 \cdot 1} = \frac{1}{2}$

Both methods yield the same result.

The value of the limit is $\frac{1}{2}$. This is also a standard limit, related to the definition of the derivative of $\sqrt{x}$ at $x=1$.

Comparing this result with the given options, the correct option is (B) 1/2.

We are asked to find the derivative of the function $f(x) = x \cos x$.

This function is a product of two functions: $u(x) = x$ and $v(x) = \cos x$.

We will use the product rule for differentiation.

The product rule states that if $f(x) = u(x)v(x)$, then the derivative $f'(x)$ is given by:

$f'(x) = u(x)v'(x) + v(x)u'(x)$

Let $u(x) = x$. Find its derivative with respect to $x$:

$u'(x) = \frac{d}{dx}(x) = 1$

Let $v(x) = \cos x$. Find its derivative with respect to $x$:

$v'(x) = \frac{d}{dx}(\cos x) = -\sin x$

Now, apply the product rule formula $f'(x) = u(x)v'(x) + v(x)u'(x)$:

$f'(x) = (x)(-\sin x) + (\cos x)(1)$

$f'(x) = -x \sin x + \cos x$

Rearranging the terms, we get:

$f'(x) = \cos x - x \sin x$

The derivative of $f(x) = x \cos x$ is $\cos x - x \sin x$.

Comparing this result with the given options, the correct option is (A) $\cos x - x \sin x$.

Let's analyze the given Assertion (A) and Reason (R).

Assertion (A): If $\lim\limits_{x \to a} f(x)$ exists, then $f(a)$ must be defined.

The definition of the limit of a function $f(x)$ as $x$ approaches $a$, $\lim\limits_{x \to a} f(x)$, describes the behavior of the function values as $x$ gets arbitrarily close to $a$, but not equal to $a$.

Consider the function $f(x) = \frac{x^2 - 4}{x - 2}$.

The limit as $x \to 2$ is $\lim\limits_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim\limits_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim\limits_{x \to 2} (x+2) = 2 + 2 = 4$.

The limit exists and is equal to 4.

However, the function $f(x)$ is not defined at $x=2$, because substituting $x=2$ results in division by zero ($f(2) = \frac{2^2 - 4}{2 - 2} = \frac{0}{0}$, which is undefined).

This is a counterexample showing that the existence of the limit at a point does not require the function to be defined at that point.

Therefore, Assertion (A) is false.

Reason (R): The existence of the limit at a point depends on the behavior of the function near the point, not necessarily at the point itself.

This statement accurately reflects the definition of a limit. The limit $\lim\limits_{x \to a} f(x)$ exists if the function values $f(x)$ approach a single finite value as $x$ approaches $a$ from both the left and the right sides, considering values of $x$ in a neighborhood of $a$ but excluding $a$ itself.

Therefore, Reason (R) is true.

We have determined that Assertion (A) is false and Reason (R) is true.

Looking at the given options:

(A) Both A and R are true and R is the correct explanation of A. (False, because A is false)

(B) Both A and R are true but R is not the correct explanation of A. (False, because A is false)

(C) A is true but R is false. (False, because A is false and R is true)

(D) A is false but R is true. (This matches our findings)

The correct option is (D) A is false but R is true.

We are asked to evaluate the limit:

$\lim\limits_{x \to 0} \frac{\log_e (1+x)}{x}$

If we substitute $x = 0$ directly into the expression, we get $\frac{\log_e (1+0)}{0} = \frac{\log_e 1}{0} = \frac{0}{0}$, which is an indeterminate form.

Method 1: Using the Definition of the Derivative

This limit is related to the definition of the derivative of the function $f(y) = \log_e y$ at a specific point.

The derivative of $f(y)$ at $y=a$ is given by $f'(a) = \lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$.

Let's consider the derivative of $f(y) = \log_e y$ at $y=1$. We know the derivative of $\log_e y$ is $f'(y) = \frac{1}{y}$. So, $f'(1) = \frac{1}{1} = 1$.

Using the limit definition at $y=1$:

$f'(1) = \lim\limits_{h \to 0} \frac{\log_e(1+h) - \log_e(1)}{h}$

Since $\log_e(1) = 0$, this becomes:

$f'(1) = \lim\limits_{h \to 0} \frac{\log_e(1+h) - 0}{h} = \lim\limits_{h \to 0} \frac{\log_e(1+h)}{h}$

Replacing the dummy variable $h$ with $x$, we get the given limit:

$\lim\limits_{x \to 0} \frac{\log_e(1+x)}{x}$

Since this limit is equal to $f'(1)$, which is 1, the value of the limit is 1.

Method 2: Using L'Hopital's Rule

Since substituting $x=0$ gives the indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule.

Let $f(x) = \log_e (1+x)$ and $g(x) = x$.

Find the derivatives of $f(x)$ and $g(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(\log_e (1+x)) = \frac{1}{1+x} \cdot \frac{d}{dx}(1+x) = \frac{1}{1+x} \cdot 1 = \frac{1}{1+x}$

$g'(x) = \frac{d}{dx}(x) = 1$

Apply L'Hopital's Rule:

$\lim\limits_{x \to 0} \frac{\log_e (1+x)}{x} = \lim\limits_{x \to 0} \frac{f'(x)}{g'(x)} = \lim\limits_{x \to 0} \frac{\frac{1}{1+x}}{1}$

$= \lim\limits_{x \to 0} \frac{1}{1+x}$

Now, substitute $x = 0$ into the simplified expression:

$= \frac{1}{1+0} = \frac{1}{1} = 1$

Both methods yield the same result. This is a standard limit formula.

The value of the limit is 1.

Comparing this result with the given options, the correct option is (B) 1.

We are asked to find the derivative of the function $f(x) = x \tan x$.

This function is a product of two functions: $u(x) = x$ and $v(x) = \tan x$.

We will use the product rule for differentiation.

The product rule states that if $f(x) = u(x)v(x)$, then the derivative $f'(x)$ is given by:

$f'(x) = u(x)v'(x) + v(x)u'(x)$

Let $u(x) = x$. Find its derivative with respect to $x$:

$u'(x) = \frac{d}{dx}(x) = 1$

Let $v(x) = \tan x$. Find its derivative with respect to $x$:

$v'(x) = \frac{d}{dx}(\tan x) = \sec^2 x$

Now, apply the product rule formula $f'(x) = u(x)v'(x) + v(x)u'(x)$:

$f'(x) = (x)(\sec^2 x) + (\tan x)(1)$

$f'(x) = x \sec^2 x + \tan x$

Rearranging the terms, we get:

$f'(x) = \tan x + x \sec^2 x$

The derivative of $f(x) = x \tan x$ is $\tan x + x \sec^2 x$.

Comparing this result with the given options, the correct option is (A) $\tan x + x \sec^2 x$.

We are asked to evaluate the limit:

$\lim\limits_{x \to 0} \frac{(1+x)^n - 1}{x}$

If we substitute $x = 0$ directly into the expression, we get $\frac{(1+0)^n - 1}{0} = \frac{1^n - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0}$, which is an indeterminate form.

Method 1: Using the Definition of the Derivative

This limit is exactly the definition of the derivative of the function $f(y) = y^n$ at $y=1$.

The derivative of $f(y) = y^n$ at $y=1$ is given by $f'(1) = \lim\limits_{h \to 0} \frac{f(1+h) - f(1)}{h}$.

Here $f(y) = y^n$, so $f(1) = 1^n = 1$ and $f(1+h) = (1+h)^n$.

$f'(1) = \lim\limits_{h \to 0} \frac{(1+h)^n - 1^n}{h} = \lim\limits_{h \to 0} \frac{(1+h)^n - 1}{h}$.

Replacing the dummy variable $h$ with $x$, we get the given limit:

$\lim\limits_{x \to 0} \frac{(1+x)^n - 1}{x}$

The derivative of $f(y) = y^n$ is $f'(y) = ny^{n-1}$.

Evaluating this derivative at $y=1$, we get $f'(1) = n(1)^{n-1} = n \cdot 1 = n$.

Therefore, the value of the limit is $n$.

Method 2: Using Binomial Expansion

Using the binomial expansion for $(1+x)^n$ for small $x$ (or for any $x$ if $n$ is a positive integer):

$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots$

Substitute this into the numerator:

$(1+x)^n - 1 = (1 + nx + \frac{n(n-1)}{2!}x^2 + \dots) - 1$

$= nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots$

Now, substitute this back into the limit expression:

$\lim\limits_{x \to 0} \frac{nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots}{x}$

For $x \neq 0$, we can divide each term in the numerator by $x$:

$\lim\limits_{x \to 0} \left(\frac{nx}{x} + \frac{\frac{n(n-1)}{2!}x^2}{x} + \frac{\frac{n(n-1)(n-2)}{3!}x^3}{x} + \dots\right)$

$= \lim\limits_{x \to 0} \left(n + \frac{n(n-1)}{2!}x + \frac{n(n-1)(n-2)}{3!}x^2 + \dots\right)$

Now, evaluate the limit as $x \to 0$. All terms containing $x$ will tend to 0:

$= n + 0 + 0 + \dots = n$

Method 3: Using L'Hopital's Rule

Since substituting $x=0$ gives the indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule.

Let $f(x) = (1+x)^n - 1$ and $g(x) = x$.

Find the derivatives of $f(x)$ and $g(x)$ with respect to $x$. Use the chain rule for $f(x)$:

$f'(x) = \frac{d}{dx}((1+x)^n - 1) = n(1+x)^{n-1} \cdot \frac{d}{dx}(1+x) - 0$

$f'(x) = n(1+x)^{n-1} \cdot (1)$

$f'(x) = n(1+x)^{n-1}$

$g'(x) = \frac{d}{dx}(x) = 1$

Apply L'Hopital's Rule:

$\lim\limits_{x \to 0} \frac{(1+x)^n - 1}{x} = \lim\limits_{x \to 0} \frac{f'(x)}{g'(x)} = \lim\limits_{x \to 0} \frac{n(1+x)^{n-1}}{1}$

$= \lim\limits_{x \to 0} n(1+x)^{n-1}$

Now, substitute $x = 0$ into the simplified expression:

$= n(1+0)^{n-1} = n(1)^{n-1} = n \cdot 1 = n$

All three methods yield the same result. This is a standard limit formula.

The value of the limit is $n$.

Comparing this result with the given options, the correct option is (A) $n$.

We are asked to find the derivative of the function $f(x) = x^3 - \sin x$.

To find the derivative of this function, we can use the difference rule for differentiation.

The difference rule states that the derivative of a difference of two functions is the difference of their derivatives:

$\frac{d}{dx}(u(x) - v(x)) = \frac{du}{dx} - \frac{dv}{dx}$

Here, $u(x) = x^3$ and $v(x) = \sin x$.

The derivative of $f(x) = x^3 - \sin x$ is:

$f'(x) = \frac{d}{dx}(x^3 - \sin x)$

$f'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(\sin x)$

Now, we differentiate each term:

1. Derivative of $x^3$:

Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$, with $n=3$:

$\frac{d}{dx}(x^3) = 3x^{3-1} = 3x^2$

2. Derivative of $\sin x$:

Using the standard differentiation formula for $\sin x$:

$\frac{d}{dx}(\sin x) = \cos x$

Substitute these derivatives back into the expression for $f'(x)$:

$f'(x) = 3x^2 - \cos x$

The derivative of $f(x) = x^3 - \sin x$ is $3x^2 - \cos x$.

Comparing this result with the given options, the correct option is (A) $3x^2 - \cos x$.

Solution:

In calculus, the derivative of a function is a fundamental concept which measures the sensitivity to change of the function's value (output value) with respect to a change in its argument (input value).

The process used to find this derivative is called differentiation.

Differentiation is one of the two core operations of calculus, the other being integration.

The question asks for the name of the process of finding the derivative.

Based on the definition, the process of finding the derivative of a function is called differentiation.

Therefore, the blank should be filled with the word "Differentiation".

The correct option is (B) Differentiation.

Solution:

We need to evaluate the limit $\lim\limits_{x \to 0} \frac{\sin x - \tan x}{x^3}$.

As $x \to 0$, $\sin x \to 0$ and $\tan x \to 0$, so the numerator $\sin x - \tan x \to 0$. The denominator $x^3 \to 0$.

This is an indeterminate form of type $\frac{0}{0}$.

We can use Taylor series expansions for $\sin x$ and $\tan x$ around $x=0$:

$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... = x - \frac{x^3}{6} + O(x^5)$

$\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + ... = x + \frac{x^3}{3} + O(x^5)$

Substitute these expansions into the expression:

$\frac{\sin x - \tan x}{x^3} = \frac{(x - \frac{x^3}{6} + O(x^5)) - (x + \frac{x^3}{3} + O(x^5))}{x^3}$

Simplify the numerator:

Numerator $= x - \frac{x^3}{6} - x - \frac{x^3}{3} + O(x^5)$

Numerator $= -\frac{x^3}{6} - \frac{x^3}{3} + O(x^5)$

Numerator $= x^3 \left(-\frac{1}{6} - \frac{1}{3}\right) + O(x^5)$

Numerator $= x^3 \left(-\frac{1}{6} - \frac{2}{6}\right) + O(x^5)$

Numerator $= x^3 \left(-\frac{3}{6}\right) + O(x^5)$

Numerator $= -\frac{x^3}{2} + O(x^5)$

Now substitute this back into the limit expression:

$\lim\limits_{x \to 0} \frac{-\frac{x^3}{2} + O(x^5)}{x^3}$

$\lim\limits_{x \to 0} \left(\frac{-\frac{x^3}{2}}{x^3} + \frac{O(x^5)}{x^3}\right)$

$\lim\limits_{x \to 0} \left(-\frac{1}{2} + O(x^2)\right)$

As $x \to 0$, the $O(x^2)$ terms go to 0.

So, the limit is $-\frac{1}{2}$.

Alternatively, we could use L'Hôpital's Rule three times, but the Taylor series method is more straightforward in this case.

The value of the limit is $-\frac{1}{2}$.

The correct option is (C) -1/2.

Solution:

We are asked to find the derivative of the function $f(x) = (ax + b)^n$.

This function is a composite function, so we will use the chain rule to find its derivative.

The chain rule states that if $f(x) = g(h(x))$, then $f'(x) = g'(h(x)) \cdot h'(x)$.

In this case, let the outer function be $g(u) = u^n$, and the inner function be $h(x) = ax + b$.

Find the derivative of the outer function $g(u)$ with respect to $u$:

$g'(u) = \frac{d}{du}(u^n) = nu^{n-1}$

Find the derivative of the inner function $h(x)$ with respect to $x$:

$h'(x) = \frac{d}{dx}(ax + b) = a \frac{d}{dx}(x) + \frac{d}{dx}(b)$

$h'(x) = a(1) + 0 = a$

Now apply the chain rule: $f'(x) = g'(h(x)) \cdot h'(x)$.

Substitute $h(x) = ax + b$ into $g'(u) = nu^{n-1}$ to get $g'(h(x)) = n(ax+b)^{n-1}$.

Multiply this by $h'(x) = a$.

$f'(x) = n(ax+b)^{n-1} \cdot a$

$f'(x) = an(ax+b)^{n-1}$

Thus, the derivative of $f(x) = (ax + b)^n$ is $an(ax+b)^{n-1}$.

The correct option is (B) $an(ax+b)^{n-1}$.

Solution:

We need to evaluate the limit $\lim\limits_{x \to 0} \frac{\sin ax}{\tan bx}$.

As $x \to 0$, the numerator $\sin ax \to \sin(0) = 0$ and the denominator $\tan bx \to \tan(0) = 0$.

This is an indeterminate form of the type $\frac{0}{0}$.

We can use the standard trigonometric limits:

$\lim\limits_{y \to 0} \frac{\sin y}{y} = 1$

$\lim\limits_{y \to 0} \frac{\tan y}{y} = 1$

We can rewrite the given expression by dividing both the numerator and the denominator by $x$ (since $x \to 0$, $x \neq 0$):

$\frac{\sin ax}{\tan bx} = \frac{\frac{\sin ax}{x}}{\frac{\tan bx}{x}}$

To apply the standard limits, we multiply and divide the terms in the numerator and denominator by appropriate constants:

$\frac{\frac{\sin ax}{x}}{\frac{\tan bx}{x}} = \frac{\frac{\sin ax}{ax} \cdot a}{\frac{\tan bx}{bx} \cdot b}$

Now, we can evaluate the limit:

$\lim\limits_{x \to 0} \frac{\frac{\sin ax}{ax} \cdot a}{\frac{\tan bx}{bx} \cdot b} = \frac{\lim\limits_{x \to 0} \left(\frac{\sin ax}{ax}\right) \cdot a}{\lim\limits_{x \to 0} \left(\frac{\tan bx}{bx}\right) \cdot b}$

As $x \to 0$, $ax \to 0$ and $bx \to 0$. Using the standard limits:

$\lim\limits_{x \to 0} \frac{\sin ax}{ax} = 1$

$\lim\limits_{x \to 0} \frac{\tan bx}{bx} = 1$

Substitute these values back into the limit expression:

$\frac{1 \cdot a}{1 \cdot b} = \frac{a}{b}$

Thus, the value of the limit is $\frac{a}{b}$.

The correct option is (A) $a/b$.

Solution:

We need to find the derivative of the function $f(x) = \sec x$ with respect to $x$.

We know that $\sec x = \frac{1}{\cos x}$.

So, we need to find the derivative of $\frac{1}{\cos x}$.

We can use the quotient rule or rewrite it as $(\cos x)^{-1}$ and use the chain rule.

Using the chain rule: Let $y = (\cos x)^{-1}$. Let $u = \cos x$. Then $y = u^{-1}$.

The derivative with respect to $u$ is $\frac{dy}{du} = -1 \cdot u^{-2} = -\frac{1}{u^2}$.

The derivative of $u = \cos x$ with respect to $x$ is $\frac{du}{dx} = -\sin x$.

By the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

$\frac{dy}{dx} = \left(-\frac{1}{u^2}\right) \cdot (-\sin x)$

Substitute $u = \cos x$ back:

$\frac{dy}{dx} = \left(-\frac{1}{(\cos x)^2}\right) \cdot (-\sin x)$

$\frac{dy}{dx} = \frac{\sin x}{\cos^2 x}$

We can rewrite $\frac{\sin x}{\cos^2 x}$ as $\frac{\sin x}{\cos x} \cdot \frac{1}{\cos x}$.

We know that $\frac{\sin x}{\cos x} = \tan x$ and $\frac{1}{\cos x} = \sec x$.

So, the derivative is $\tan x \cdot \sec x$ or $\sec x \tan x$.

The derivative of $f(x) = \sec x$ is $\sec x \tan x$.

The correct option is (B) $\sec x \tan x$.

Solution:

Let's evaluate the given Assertion (A) and Reason (R).

Assertion (A): The derivative of a product of two functions $u(x)$ and $v(x)$ is $u'v' + uv$.

The correct product rule for differentiation states that $\frac{d}{dx}(u \cdot v) = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx}$, which can be written as $uv' + vu'$.

The formula given in Assertion (A), $u'v' + uv$, is incorrect.

Therefore, Assertion (A) is false.

Reason (R): The correct product rule is $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$.

This statement provides the standard and correct formula for the product rule in differentiation.

Therefore, Reason (R) is true.

Based on the evaluation:

Assertion (A) is false.

Reason (R) is true.

Looking at the options:

(A) Both A and R are true... (Incorrect, A is false)

(B) Both A and R are true... (Incorrect, A is false)

(C) A is true but R is false. (Incorrect, A is false and R is true)

(D) A is false but R is true. (Correct)

The correct option is (D) A is false but R is true.

Solution:

We need to evaluate the limit $\lim\limits_{x \to 0} \frac{e^{ax} - 1}{bx}$.

As $x \to 0$, the numerator $e^{ax} - 1 \to e^{a \cdot 0} - 1 = e^0 - 1 = 1 - 1 = 0$.

As $x \to 0$, the denominator $bx \to b \cdot 0 = 0$ (since $b \neq 0$).

This is an indeterminate form of the type $\frac{0}{0}$.

We can use the standard limit $\lim\limits_{y \to 0} \frac{e^y - 1}{y} = 1$.

To apply this standard limit, we manipulate the given expression:

$\frac{e^{ax} - 1}{bx} = \frac{1}{b} \cdot \frac{e^{ax} - 1}{x}$

Now, we need to make the denominator of the fraction match the exponent in the numerator. We can multiply and divide the second term by $a$:

$\frac{1}{b} \cdot \frac{e^{ax} - 1}{x} = \frac{1}{b} \cdot \frac{e^{ax} - 1}{ax} \cdot a$

$\frac{e^{ax} - 1}{bx} = \frac{a}{b} \cdot \frac{e^{ax} - 1}{ax}$

Now, take the limit as $x \to 0$:

$\lim\limits_{x \to 0} \frac{e^{ax} - 1}{bx} = \lim\limits_{x \to 0} \left(\frac{a}{b} \cdot \frac{e^{ax} - 1}{ax}\right)$

Using the properties of limits, we can write this as:

$\lim\limits_{x \to 0} \frac{a}{b} \cdot \lim\limits_{x \to 0} \frac{e^{ax} - 1}{ax}$

The first part is a constant, $\frac{a}{b}$.

For the second part, let $y = ax$. As $x \to 0$, $y \to a \cdot 0 = 0$. So, $\lim\limits_{x \to 0} \frac{e^{ax} - 1}{ax} = \lim\limits_{y \to 0} \frac{e^y - 1}{y}$.

From the standard limit, we know $\lim\limits_{y \to 0} \frac{e^y - 1}{y} = 1$.

Substitute this back into the limit expression:

$\frac{a}{b} \cdot 1 = \frac{a}{b}$

Thus, the value of the limit is $\frac{a}{b}$.

The correct option is (A) $a/b$.

Solution:

We need to find the derivative of the function $f(x) = \cot x$ with respect to $x$.

We can express $\cot x$ in terms of $\sin x$ and $\cos x$ as $\cot x = \frac{\cos x}{\sin x}$.

We will use the quotient rule for differentiation, which states that if $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}$.

Let $u(x) = \cos x$ and $v(x) = \sin x$.

Find the derivative of $u(x)$: $u'(x) = \frac{d}{dx}(\cos x) = -\sin x$.

Find the derivative of $v(x)$: $v'(x) = \frac{d}{dx}(\sin x) = \cos x$.

Apply the quotient rule:

$f'(x) = \frac{(\sin x)(-\sin x) - (\cos x)(\cos x)}{(\sin x)^2}$

$f'(x) = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x}$

Factor out $-1$ from the numerator:

$f'(x) = \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x}

Using the fundamental trigonometric identity $\sin^2 x + \cos^2 x = 1$:

$f'(x) = \frac{-(1)}{\sin^2 x}

$f'(x) = -\frac{1}{\sin^2 x}

Using the reciprocal identity $\text{cosec } x = \frac{1}{\sin x}$:

$f'(x) = -(\frac{1}{\sin x})^2 = -\text{cosec}^2 x$.

Thus, the derivative of $f(x) = \cot x$ is $-\text{cosec}^2 x$.

The correct option is (C) $-\text{cosec}^2 x$.

Solution:

We are asked to identify the limit of an exponential function from the given options.

Let's examine each option:

(A) $\lim\limits_{x \to 0} \frac{\sin x}{x}$: This is a standard limit involving trigonometric functions. Its value is 1.

(B) $\lim\limits_{x \to a} \frac{x^n - a^n}{x - a}$: This is related to the derivative of a power function $x^n$ or a standard limit for algebraic functions. Its value is $na^{n-1}$.

(C) $\lim\limits_{x \to 0} \frac{e^x - 1}{x}$: This is a standard limit involving the exponential function $e^x$. Its value is 1.

(D) $\lim\limits_{x \to \infty} \frac{1}{x}$: This is a limit involving a simple rational function as the variable approaches infinity. Its value is 0.

The limit that directly involves an exponential function ($e^x$ in this case) is option (C).

Therefore, the limit of an exponential function among the given options is $\lim\limits_{x \to 0} \frac{e^x - 1}{x}$.

The correct option is (C) $\lim\limits_{x \to 0} \frac{e^x - 1}{x}$.

Solution:

We need to find the derivative of the function $f(x) = x^2 \tan x$.

This function is a product of two functions: $u(x) = x^2$ and $v(x) = \tan x$.

We will use the product rule for differentiation, which states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$, or $\frac{d}{dx}(uv) = \frac{du}{dx}v + u\frac{dv}{dx}$.

Let $u(x) = x^2$.

The derivative of $u(x)$ with respect to $x$ is $u'(x) = \frac{d}{dx}(x^2) = 2x$.

Let $v(x) = \tan x$.

The derivative of $v(x)$ with respect to $x$ is $v'(x) = \frac{d}{dx}(\tan x) = \sec^2 x$.

Now, apply the product rule:

$f'(x) = u'(x)v(x) + u(x)v'(x)$

$f'(x) = (2x)(\tan x) + (x^2)(\sec^2 x)$

$f'(x) = 2x \tan x + x^2 \sec^2 x$

Thus, the derivative of $f(x) = x^2 \tan x$ is $2x \tan x + x^2 \sec^2 x$.

The correct option is (A) $2x \tan x + x^2 \sec^2 x$.

Solution:

We need to evaluate each given limit and determine which one results in a value that is not 1, 0, $e$, or a finite constant.

Let's evaluate each option:

(A) $\lim\limits_{x \to 0} \frac{\sin x}{x}$

This is a standard trigonometric limit.

$\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$

The value is 1, which is in the set of excluded values (1, 0, $e$, or a finite constant).

(B) $\lim\limits_{x \to \infty} \frac{x^2+1}{x}$

As $x \to \infty$, we can analyze the highest power terms. The highest power in the numerator is $x^2$ and in the denominator is $x$.

Divide both numerator and denominator by $x$:

$\lim\limits_{x \to \infty} \frac{\frac{x^2}{x}+\frac{1}{x}}{\frac{x}{x}} = \lim\limits_{x \to \infty} \frac{x+\frac{1}{x}}{1} = \lim\limits_{x \to \infty} (x + \frac{1}{x})$

As $x \to \infty$, $x \to \infty$ and $\frac{1}{x} \to 0$.

So, the limit is $\infty + 0 = \infty$.

The value is $\infty$, which is not a finite constant, nor is it 1, 0, or $e$.

(C) $\lim\limits_{x \to 1} \frac{x-1}{x+1}$

This function is continuous at $x=1$. We can evaluate the limit by direct substitution:

$\lim\limits_{x \to 1} \frac{x-1}{x+1} = \frac{1-1}{1+1} = \frac{0}{2} = 0$

The value is 0, which is in the set of excluded values (1, 0, $e$, or a finite constant).

(D) $\lim\limits_{x \to 0} (1+x)^{1/x}$

This is a fundamental limit definition of the mathematical constant $e$.

$\lim\limits_{x \to 0} (1+x)^{1/x} = e$

The value is $e$, which is in the set of excluded values (1, 0, $e$, or a finite constant).

Comparing the results, only option (B) yields a limit value ($\infty$) that is not 1, 0, $e$, or a finite constant.

The correct option is (B) $\lim\limits_{x \to \infty} \frac{x^2+1}{x}$.

Solution:

We need to evaluate the limit $\lim\limits_{x \to 2} \frac{x^3 - 8}{x^2 - 4}$.

As $x \to 2$, the numerator $x^3 - 8 \to 2^3 - 8 = 8 - 8 = 0$.

As $x \to 2$, the denominator $x^2 - 4 \to 2^2 - 4 = 4 - 4 = 0$.

This is an indeterminate form of the type $\frac{0}{0}$.

We can factor the numerator and the denominator.

The numerator is a difference of cubes: $x^3 - 8 = x^3 - 2^3 = (x - 2)(x^2 + 2x + 2^2) = (x - 2)(x^2 + 2x + 4)$.

The denominator is a difference of squares: $x^2 - 4 = x^2 - 2^2 = (x - 2)(x + 2)$.

Substitute the factored forms into the limit expression:

$\lim\limits_{x \to 2} \frac{(x - 2)(x^2 + 2x + 4)}{(x - 2)(x + 2)}$

Since $x \to 2$, $x \neq 2$, so $x - 2 \neq 0$. We can cancel the $(x - 2)$ term from the numerator and the denominator:

$\lim\limits_{x \to 2} \frac{\cancel{(x - 2)}(x^2 + 2x + 4)}{\cancel{(x - 2)}(x + 2)} = \lim\limits_{x \to 2} \frac{x^2 + 2x + 4}{x + 2}$

Now, the function $\frac{x^2 + 2x + 4}{x + 2}$ is continuous at $x = 2$ because the denominator $x+2$ is not zero at $x=2$ ($2+2=4$).

So, we can evaluate the limit by direct substitution:

$\lim\limits_{x \to 2} \frac{x^2 + 2x + 4}{x + 2} = \frac{2^2 + 2(2) + 4}{2 + 2} = \frac{4 + 4 + 4}{4} = \frac{12}{4} = 3$

Thus, the value of the limit is 3.

The correct option is (C) 3.

Solution:

We need to find the derivative of the function $f(x) = x^4$ with respect to $x$.

We will use the power rule for differentiation, which states that for any real number $n$, the derivative of $x^n$ is $\frac{d}{dx}(x^n) = nx^{n-1}$.

In our function $f(x) = x^4$, the base is $x$ and the exponent is $n=4$.

Applying the power rule, we have:

$f'(x) = \frac{d}{dx}(x^4) = 4 \cdot x^{4-1}$

$f'(x) = 4x^3$

Thus, the derivative of $f(x) = x^4$ is $4x^3$.

The correct option is (A) $4x^3$.

Solution:

The question describes a property of differentiation regarding the derivative of a sum or difference of functions.

Let $u(x)$ and $v(x)$ be two differentiable functions.

The property stated is that the derivative of $(u(x) + v(x))$ is $\frac{du}{dx} + \frac{dv}{dx}$, and the derivative of $(u(x) - v(x))$ is $\frac{du}{dx} - \frac{dv}{dx}$.

This can be formally written as:

$\frac{d}{dx}[u(x) + v(x)] = \frac{d}{dx}[u(x)] + \frac{d}{dx}[v(x)]$

$\frac{d}{dx}[u(x) - v(x)] = \frac{d}{dx}[u(x)] - \frac{d}{dx}[v(x)]$

Let's examine the given options:

(A) Product Rule: This rule is used for finding the derivative of a product of two functions: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$.

(B) Quotient Rule: This rule is used for finding the derivative of a quotient of two functions: $\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.

(C) Sum/Difference Rule: This rule states that the derivative of a sum or difference of functions is the sum or difference of their derivatives. This matches the description in the question.

(D) Chain Rule: This rule is used for finding the derivative of a composite function: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$.

The rule that describes the derivative of a sum or difference being the sum or difference of the derivatives is the Sum/Difference Rule.

The correct option is (C) Sum/Difference Rule.

Solution:

The question shows a method for evaluating the limit $\lim\limits_{x \to 0} \frac{\sin ax}{\sin bx}$.

The method involves rewriting the expression as:

$\frac{\sin ax}{\sin bx} = \frac{\sin ax}{ax} \cdot ax \cdot \frac{1}{\sin bx} = \frac{\sin ax}{ax} \cdot ax \cdot \frac{1}{\frac{\sin bx}{bx} \cdot bx} = \frac{\sin ax}{ax} \cdot \frac{ax}{\frac{\sin bx}{bx} \cdot bx}$

This is equivalent to the form given in the question: $\frac{\sin ax}{ax} \cdot \frac{bx}{\sin bx} \cdot \frac{ax}{bx}$.

We can simplify the $\frac{ax}{bx}$ term (assuming $b \neq 0$ and $x \neq 0$ as $x \to 0$): $\frac{a}{b}$.

So the expression becomes: $\frac{\sin ax}{ax} \cdot \frac{bx}{\sin bx} \cdot \frac{a}{b}$.

Taking the limit as $x \to 0$:

$\lim\limits_{x \to 0} \left(\frac{\sin ax}{ax} \cdot \frac{bx}{\sin bx} \cdot \frac{a}{b}\right)$

This limit is evaluated by using the property of limits that the limit of a product is the product of the limits, provided the individual limits exist.

We use the standard trigonometric limit $\lim\limits_{y \to 0} \frac{\sin y}{y} = 1$.

As $x \to 0$, $ax \to 0$, so $\lim\limits_{x \to 0} \frac{\sin ax}{ax} = 1$.

Similarly, as $x \to 0$, $bx \to 0$, so $\lim\limits_{x \to 0} \frac{\sin bx}{bx} = 1$.

Therefore, $\lim\limits_{x \to 0} \frac{bx}{\sin bx} = \frac{1}{\lim\limits_{x \to 0} \frac{\sin bx}{bx}} = \frac{1}{1} = 1$.

Substituting these values into the limit expression:

$\lim\limits_{x \to 0} \left(\frac{\sin ax}{ax}\right) \cdot \lim\limits_{x \to 0} \left(\frac{bx}{\sin bx}\right) \cdot \lim\limits_{x \to 0} \left(\frac{a}{b}\right)$

$= 1 \cdot 1 \cdot \frac{a}{b} = \frac{a}{b}$.

The key technique used in this method is the application of the standard trigonometric limit $\lim\limits_{y \to 0} \frac{\sin y}{y} = 1$ (and its reciprocal). The expression was manipulated to make these standard forms appear.

Let's consider the options:

(A) Direct Substitution: Leads to $\frac{0}{0}$, an indeterminate form, so not used directly.

(B) Factorization: Not the primary technique here, though factors were separated.

(C) Standard Limits: This is the core method used by transforming the expression into forms whose limits are known standard values.

(D) L'Hopital's Rule: While this is a valid technique for $\frac{0}{0}$ forms, the method shown in the question explicitly uses manipulation to apply standard limits, not differentiation of the numerator and denominator.

The example provided demonstrates the use of standard limits.

The correct option is (C) Standard Limits.

Solution:

We need to find the derivative of the function $f(x) = x \log_e x$.

This function is a product of two functions: $u(x) = x$ and $v(x) = \log_e x$.

We will use the product rule for differentiation, which states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$, or $\frac{d}{dx}(uv) = \frac{du}{dx}v + u\frac{dv}{dx}$.

Let $u(x) = x$.

The derivative of $u(x)$ with respect to $x$ is $u'(x) = \frac{d}{dx}(x) = 1$.

Let $v(x) = \log_e x$.

The derivative of $v(x)$ with respect to $x$ is $v'(x) = \frac{d}{dx}(\log_e x) = \frac{1}{x}$.

Now, apply the product rule:

$f'(x) = u'(x)v(x) + u(x)v'(x)$

Substitute the functions and their derivatives:

$f'(x) = (1)(\log_e x) + (x)\left(\frac{1}{x}\right)$

Simplify the expression:

$f'(x) = \log_e x + \frac{x}{x}$

Since $x$ is in the domain of $\log_e x$, we know $x \neq 0$. So $\frac{x}{x} = 1$.

$f'(x) = \log_e x + 1$

Thus, the derivative of $f(x) = x \log_e x$ is $\log_e x + 1$.

The correct option is (A) $\log_e x + 1$.

Solution:

We are asked to identify the power rule for differentiation from the given options.

Let's examine each option and the rule it represents:

(A) $\frac{d}{dx}(c) = 0$: This is the rule for the derivative of a constant, where $c$ is a constant.

(B) $\frac{d}{dx}(x) = 1$: This is the rule for the derivative of $x$ with respect to $x$. It is a special case of the power rule where $n=1$, as $\frac{d}{dx}(x^1) = 1 \cdot x^{1-1} = 1 \cdot x^0 = 1 \cdot 1 = 1$.

(C) $\frac{d}{dx}(x^n) = nx^{n-1}$: This is the general formula for the derivative of $x$ raised to a power $n$, where $n$ is any real number. This is the definition of the power rule for differentiation.

(D) $\frac{d}{dx}(e^x) = e^x$: This is the rule for the derivative of the natural exponential function $e^x$.

The statement $\frac{d}{dx}(x^n) = nx^{n-1}$ is the correct formulation of the power rule for differentiation.

The correct option is (C) $\frac{d}{dx}(x^n) = nx^{n-1}$.

Solution:

We need to evaluate the limit $\lim\limits_{x \to \infty} \frac{5x^3 - 2x + 1}{2x^4 + 3x^2 - 5}$.

This is a limit of a rational function as $x$ approaches infinity.

A standard method for evaluating such limits is to divide both the numerator and the denominator by the highest power of $x$ present in the denominator.

The highest power of $x$ in the denominator $(2x^4 + 3x^2 - 5)$ is $x^4$.

Divide each term in the numerator and the denominator by $x^4$:

$\frac{5x^3 - 2x + 1}{2x^4 + 3x^2 - 5} = \frac{\frac{5x^3}{x^4} - \frac{2x}{x^4} + \frac{1}{x^4}}{\frac{2x^4}{x^4} + \frac{3x^2}{x^4} - \frac{5}{x^4}}$

Simplify each term:

Numerator terms: $\frac{5x^3}{x^4} = \frac{5}{x}$, $\frac{2x}{x^4} = \frac{2}{x^3}$, $\frac{1}{x^4}$

Denominator terms: $\frac{2x^4}{x^4} = 2$, $\frac{3x^2}{x^4} = \frac{3}{x^2}$, $\frac{5}{x^4}$

The expression becomes:

$\frac{\frac{5}{x} - \frac{2}{x^3} + \frac{1}{x^4}}{2 + \frac{3}{x^2} - \frac{5}{x^4}}$

Now, evaluate the limit as $x \to \infty$. Remember that for any positive integer $k$, $\lim\limits_{x \to \infty} \frac{c}{x^k} = 0$ for any constant $c$.

$\lim\limits_{x \to \infty} \frac{\frac{5}{x} - \frac{2}{x^3} + \frac{1}{x^4}}{2 + \frac{3}{x^2} - \frac{5}{x^4}} = \frac{\lim\limits_{x \to \infty} \frac{5}{x} - \lim\limits_{x \to \infty} \frac{2}{x^3} + \lim\limits_{x \to \infty} \frac{1}{x^4}}{\lim\limits_{x \to \infty} 2 + \lim\limits_{x \to \infty} \frac{3}{x^2} - \lim\limits_{x \to \infty} \frac{5}{x^4}}$

Applying the limit rule $\lim\limits_{x \to \infty} \frac{c}{x^k} = 0$ and $\lim\limits_{x \to \infty} c = c$:

$= \frac{0 - 0 + 0}{2 + 0 - 0} = \frac{0}{2} = 0$

Alternatively, we can compare the degrees of the numerator and the denominator. The degree of the numerator ($3$) is less than the degree of the denominator ($4$).

For a rational function $P(x)/Q(x)$, if the degree of $P(x)$ is less than the degree of $Q(x)$, then $\lim\limits_{x \to \infty} \frac{P(x)}{Q(x)} = 0$.

Using this rule, the limit is 0.

The value of the limit is 0.

The correct option is (C) 0.

Solution:

We need to find the derivative of the function $f(x) = (x^2 + 5)^3$ with respect to $x$.

This is a composite function, so we will use the chain rule for differentiation.

The chain rule states that if $f(x) = g(h(x))$, then $f'(x) = g'(h(x)) \cdot h'(x)$.

In this function, the outer function is a power function, and the inner function is a polynomial.

Let $h(x) = x^2 + 5$ be the inner function.

Let $g(u) = u^3$ be the outer function, where $u = h(x)$.

First, find the derivative of the outer function $g(u)$ with respect to $u$:

$g'(u) = \frac{d}{du}(u^3)$

Using the power rule, $g'(u) = 3u^{3-1} = 3u^2$.

Next, find the derivative of the inner function $h(x)$ with respect to $x$:

$h'(x) = \frac{d}{dx}(x^2 + 5)$

Using the sum rule and power rule, $h'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(5) = 2x^1 + 0 = 2x$.

Now, apply the chain rule $f'(x) = g'(h(x)) \cdot h'(x)$.

Substitute $u = h(x) = x^2 + 5$ into $g'(u)$: $g'(h(x)) = 3(x^2 + 5)^2$.

Multiply this by $h'(x) = 2x$:

$f'(x) = 3(x^2 + 5)^2 \cdot (2x)$

Rearrange and simplify the expression:

$f'(x) = 2x \cdot 3(x^2 + 5)^2$

$f'(x) = 6x(x^2 + 5)^2$

Thus, the derivative of $f(x) = (x^2 + 5)^3$ is $6x(x^2 + 5)^2$.

Comparing this with the given options, we see that option (C) matches the simplified derivative.

Option (B) is an equivalent form before the final multiplication of constants and variables outside the parenthesis.

The correct option is (C) $6x(x^2+5)^2$.

Solution:

We need to evaluate the limit $\lim\limits_{x \to 0} \frac{e^{2x} - 1}{x}$.

As $x \to 0$, the numerator $e^{2x} - 1 \to e^{2 \cdot 0} - 1 = e^0 - 1 = 1 - 1 = 0$.

As $x \to 0$, the denominator $x \to 0$.

This is an indeterminate form of the type $\frac{0}{0}$.

We can use the standard limit $\lim\limits_{y \to 0} \frac{e^y - 1}{y} = 1$.

To apply this standard limit, we manipulate the given expression. We need the denominator to match the exponent in the numerator, which is $2x$.

We can multiply the denominator by 2 and also multiply the entire fraction by 2 (which is equivalent to multiplying the numerator by 2 and the denominator by 2):

$\frac{e^{2x} - 1}{x} = \frac{e^{2x} - 1}{x} \cdot \frac{2}{2} = 2 \cdot \frac{e^{2x} - 1}{2x}$

Now, take the limit as $x \to 0$:

$\lim\limits_{x \to 0} \left(2 \cdot \frac{e^{2x} - 1}{2x}\right)$

Using the property of limits $\lim\limits_{x \to c} k f(x) = k \lim\limits_{x \to c} f(x)$:

$= 2 \cdot \lim\limits_{x \to 0} \frac{e^{2x} - 1}{2x}$

Let $y = 2x$. As $x \to 0$, the value of $y = 2x$ also approaches 0. So, we can rewrite the limit in terms of $y$:

$\lim\limits_{x \to 0} \frac{e^{2x} - 1}{2x} = \lim\limits_{y \to 0} \frac{e^y - 1}{y}$

From the standard limit, we know $\lim\limits_{y \to 0} \frac{e^y - 1}{y} = 1$.

Substitute this value back into our limit expression:

$2 \cdot \lim\limits_{y \to 0} \frac{e^y - 1}{y} = 2 \cdot 1 = 2$

Thus, the value of the limit is 2.

The correct option is (B) 2.

Solution:

Let's evaluate the given Assertion (A) and Reason (R).

Assertion (A): The limit of $\frac{1}{x}$ as $x \to 0$ does not exist.

To determine if the limit $\lim\limits_{x \to 0} \frac{1}{x}$ exists, we consider the left-hand limit and the right-hand limit as $x$ approaches 0.

The limit exists if and only if both one-sided limits exist and are equal to a finite value.

Reason (R): The left-hand limit ($\lim\limits_{x \to 0^-} \frac{1}{x} = -\infty$) and the right-hand limit ($\lim\limits_{x \to 0^+} \frac{1}{x} = \infty$) are not equal.

Let's calculate the one-sided limits:

As $x$ approaches 0 from the right side ($x \to 0^+$), $x$ is a small positive number. Therefore, $\frac{1}{x}$ is a large positive number.

$\lim\limits_{x \to 0^+} \frac{1}{x} = \infty$

As $x$ approaches 0 from the left side ($x \to 0^-$), $x$ is a small negative number. Therefore, $\frac{1}{x}$ is a large negative number.

$\lim\limits_{x \to 0^-} \frac{1}{x} = -\infty$

The Reason correctly states that the left-hand limit is $-\infty$ and the right-hand limit is $\infty$. It also correctly states that these limits are not equal.

Therefore, Reason (R) is true.

Since the left-hand limit ($\lim\limits_{x \to 0^-} \frac{1}{x} = -\infty$) and the right-hand limit ($\lim\limits_{x \to 0^+} \frac{1}{x} = \infty$) are not equal (and are infinite), the overall limit $\lim\limits_{x \to 0} \frac{1}{x}$ does not exist.

Therefore, Assertion (A) is true.

Both Assertion (A) and Reason (R) are true. Furthermore, Reason (R) provides the correct justification (the inequality of the one-sided limits) for why the overall limit does not exist, as stated in Assertion (A).

Thus, Reason (R) is the correct explanation of Assertion (A).

The correct option is (A) Both A and R are true and R is the correct explanation of A.

Solution:

Given: The function $f(x) = \frac{x^2}{x+1}$.

To Find: The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}[f(x)]$.

The function $f(x)$ is a quotient of two functions. Let $u(x) = x^2$ and $v(x) = x+1$.

We will use the quotient rule for differentiation, which states that if $f(x) = \frac{u(x)}{v(x)}$, then its derivative is given by:

$f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}$

First, find the derivatives of $u(x)$ and $v(x)$ with respect to $x$:

$u'(x) = \frac{d}{dx}(x^2)$

Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:

$u'(x) = 2x^{2-1} = 2x$

$v'(x) = \frac{d}{dx}(x+1)$

Using the sum rule and derivative of a constant and $x$:

$v'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(1) = 1 + 0 = 1$

Now, apply the quotient rule formula $f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}$:

$f'(x) = \frac{(x+1)(2x) - (x^2)(1)}{(x+1)^2}$

Simplify the numerator:

Numerator $= (x+1)(2x) - (x^2)(1)$

Numerator $= (2x \cdot x + 2x \cdot 1) - x^2$

Numerator $= 2x^2 + 2x - x^2$

Numerator $= (2x^2 - x^2) + 2x$

Numerator $= x^2 + 2x$

Substitute the simplified numerator back into the derivative expression:

$f'(x) = \frac{x^2 + 2x}{(x+1)^2}$

Comparing this result with the given options, we see that option (A) shows the steps of applying the quotient rule and simplification, arriving at this result. Option (B) also provides the final simplified result. Given the form of option (A), it demonstrates the correct application of the rule.

The correct option is (A) $\frac{(x+1)(2x) - (x^2)(1)}{(x+1)^2} = \frac{2x^2+2x-x^2}{(x+1)^2} = \frac{x^2+2x}{(x+1)^2}$.

Solution:

We need to evaluate the limit $\lim\limits_{x \to 0} \frac{1 - \cos (ax)}{x^2}$.

As $x \to 0$, the numerator $1 - \cos(ax) \to 1 - \cos(0) = 1 - 1 = 0$.

As $x \to 0$, the denominator $x^2 \to 0^2 = 0$.

This is an indeterminate form of the type $\frac{0}{0}$.

We can use the standard trigonometric limit $\lim\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}$.

To apply this standard limit, we need the denominator to be the square of the argument of the cosine function, which is $ax$. The square of the argument is $(ax)^2 = a^2 x^2$.

We can multiply and divide the expression by $a^2$:

$\frac{1 - \cos (ax)}{x^2} = \frac{1 - \cos (ax)}{x^2} \cdot \frac{a^2}{a^2} = a^2 \cdot \frac{1 - \cos (ax)}{a^2 x^2} = a^2 \cdot \frac{1 - \cos (ax)}{(ax)^2}$

Now, evaluate the limit as $x \to 0$:

$\lim\limits_{x \to 0} \left(a^2 \cdot \frac{1 - \cos (ax)}{(ax)^2}\right)$

Using the property that the limit of a constant times a function is the constant times the limit of the function:

$= a^2 \cdot \lim\limits_{x \to 0} \frac{1 - \cos (ax)}{(ax)^2}$

Let $y = ax$. As $x \to 0$, $y = ax \to a \cdot 0 = 0$.

So, the limit becomes:

$a^2 \cdot \lim\limits_{y \to 0} \frac{1 - \cos y}{y^2}$

Using the standard limit $\lim\limits_{y \to 0} \frac{1 - \cos y}{y^2} = \frac{1}{2}$:

$= a^2 \cdot \frac{1}{2} = \frac{a^2}{2}$

Thus, the value of the limit is $\frac{a^2}{2}$.

The correct option is (A) $a^2/2$.

Solution:

We need to find the derivative of the function $f(x) = x \cdot e^x$ with respect to $x$.

This function is a product of two functions: $u(x) = x$ and $v(x) = e^x$.

We will use the product rule for differentiation, which states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$, or $\frac{d}{dx}(uv) = \frac{du}{dx}v + u\frac{dv}{dx}$.

Let $u(x) = x$.

The derivative of $u(x)$ with respect to $x$ is $u'(x) = \frac{d}{dx}(x) = 1$.

Let $v(x) = e^x$.

The derivative of $v(x)$ with respect to $x$ is $v'(x) = \frac{d}{dx}(e^x) = e^x$.

Now, apply the product rule:

$f'(x) = u'(x)v(x) + u(x)v'(x)$

Substitute the functions and their derivatives:

$f'(x) = (1)(e^x) + (x)(e^x)$

Simplify the expression:

$f'(x) = e^x + xe^x$

Thus, the derivative of $f(x) = x \cdot e^x$ is $e^x + xe^x$.

We can also factor out $e^x$: $f'(x) = e^x(1 + x)$.

Comparing the result $e^x + xe^x$ with the given options, we see that option (C) matches this result.

The correct option is (C) $e^x + xe^x$.

Solution:

We are asked to identify the correct quotient rule for differentiation from the given options.

The quotient rule is used to find the derivative of a function that is the ratio of two differentiable functions, say $u(x)$ and $v(x)$.

The correct formula for the derivative of $\frac{u(x)}{v(x)}$ with respect to $x$ is given by:

$\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{v(x) \cdot \frac{d}{dx}[u(x)] - u(x) \cdot \frac{d}{dx}[v(x)]}{[v(x)]^2}$

Using prime notation for derivatives, this can be written as:

$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2}$

Let's examine the given options:

(A) $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$: This is the product rule, not the quotient rule.

(B) $\frac{d}{dx}(\frac{u}{v}) = \frac{u\frac{dv}{dx} - v\frac{du}{dx}}{v^2}$: The terms in the numerator are in the incorrect order; it should be $v \cdot u'$ first, then subtract $u \cdot v'$. Also, the signs are incorrect relative to the standard formula.

(C) $\frac{d}{dx}(\frac{u}{v}) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$: This matches the standard formula for the quotient rule: $\frac{v \cdot u' - u \cdot v'}{v^2}$.

(D) $\frac{d}{dx}(\frac{u}{v}) = \frac{v\frac{du}{dx} + u\frac{dv}{dx}}{v^2}$: The numerator uses an addition sign instead of a subtraction sign.

Therefore, option (C) correctly represents the quotient rule for differentiation.

The correct option is (C) $\frac{d}{dx}(\frac{u}{v}) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$.

Solution:

We need to evaluate the limit $\lim\limits_{x \to 1} \frac{x^4 - 1}{x - 1}$.

As $x \to 1$, the numerator $x^4 - 1 \to 1^4 - 1 = 1 - 1 = 0$.

As $x \to 1$, the denominator $x - 1 \to 1 - 1 = 0$.

This is an indeterminate form of the type $\frac{0}{0}$.

We can factor the numerator using the difference of powers formula, $a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + ... + ab^{n-2} + b^{n-1})$. For $n=4$, $a=x$, $b=1$:
$x^4 - 1^4 = (x - 1)(x^3 \cdot 1^0 + x^2 \cdot 1^1 + x^1 \cdot 1^2 + x^0 \cdot 1^3)$

$x^4 - 1 = (x - 1)(x^3 + x^2 + x + 1)$.

Substitute the factored form into the limit expression:

$\lim\limits_{x \to 1} \frac{(x - 1)(x^3 + x^2 + x + 1)}{x - 1}$

Since $x \to 1$, $x \neq 1$, so $x - 1 \neq 0$. We can cancel the $(x - 1)$ term from the numerator and the denominator:

$\lim\limits_{x \to 1} \frac{\cancel{(x - 1)}(x^3 + x^2 + x + 1)}{\cancel{(x - 1)}} = \lim\limits_{x \to 1} (x^3 + x^2 + x + 1)$

Now, the function $(x^3 + x^2 + x + 1)$ is a polynomial, which is continuous everywhere. We can evaluate the limit by direct substitution at $x=1$:

$\lim\limits_{x \to 1} (x^3 + x^2 + x + 1) = (1)^3 + (1)^2 + (1) + 1 = 1 + 1 + 1 + 1 = 4$

Alternatively, this limit can be recognized as the definition of the derivative of $f(x) = x^4$ at $x = 1$, which is $f'(1)$. The derivative of $f(x) = x^4$ is $f'(x) = 4x^3$. Evaluating at $x=1$, we get $f'(1) = 4(1)^3 = 4$.

Thus, the value of the limit is 4.

The correct option is (C) 4.

Solution:

We need to find the derivative of the function $f(x) = \cos (2x)$ with respect to $x$.

This is a composite function, so we use the chain rule for differentiation.

The chain rule states that if $y = g(u)$ and $u = h(x)$, then the derivative of $y$ with respect to $x$ is $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

In our function $f(x) = \cos(2x)$, let the inner function be $u = 2x$ and the outer function be $f(x) = \cos(u)$.

Find the derivative of the outer function $f(x) = \cos(u)$ with respect to $u$:

$\frac{d}{du}(\cos u) = -\sin u$

Find the derivative of the inner function $u = 2x$ with respect to $x$:

$\frac{d}{dx}(2x)$

Using the constant multiple rule $\frac{d}{dx}(c \cdot g(x)) = c \cdot \frac{d}{dx}(g(x))$ and the derivative of $x$ $\frac{d}{dx}(x) = 1$:

$\frac{d}{dx}(2x) = 2 \cdot \frac{d}{dx}(x) = 2 \cdot 1 = 2$

Now, apply the chain rule: $f'(x) = \frac{d}{dx}(f(x)) = \frac{d}{du}(\cos u) \cdot \frac{du}{dx}$.

Substitute the results from the previous steps:

$f'(x) = (-\sin u) \cdot (2)$

Substitute $u = 2x$ back into the expression:

$f'(x) = (-\sin (2x)) \cdot 2$

$f'(x) = -2 \sin(2x)$

Thus, the derivative of $f(x) = \cos (2x)$ is $-2 \sin(2x)$.

The correct option is (B) $-2 \sin(2x)$.

Solution:

We need to evaluate the limit $\lim\limits_{x \to 0} \frac{(1+ax)^n - 1}{x}$.

As $x \to 0$, the numerator $(1+ax)^n - 1 \to (1+a \cdot 0)^n - 1 = 1^n - 1 = 1 - 1 = 0$.

As $x \to 0$, the denominator $x \to 0$.

This is an indeterminate form of the type $\frac{0}{0}$.

This limit is in the form of the definition of the derivative of a function at a point.

Recall the definition of the derivative of a function $f(x)$ at $x=c$:

$f'(c) = \lim\limits_{x \to c} \frac{f(x) - f(c)}{x - c}$

Let's consider the function $f(x) = (1+ax)^n$.

We want to evaluate the limit as $x \to 0$. Let's compare the given limit to the definition of the derivative at $c=0$.

If $f(x) = (1+ax)^n$, then $f(0) = (1+a \cdot 0)^n = 1^n = 1$.

The given limit is $\lim\limits_{x \to 0} \frac{(1+ax)^n - 1}{x} = \lim\limits_{x \to 0} \frac{(1+ax)^n - (1+a \cdot 0)^n}{x - 0} = \lim\limits_{x \to 0} \frac{f(x) - f(0)}{x - 0}$.

This limit is equal to the derivative of $f(x) = (1+ax)^n$ evaluated at $x=0$, i.e., $f'(0)$.

Let's find the derivative of $f(x) = (1+ax)^n$ using the chain rule.

Let $u = 1+ax$. Then $f(x) = u^n$.

$\frac{df}{du} = \frac{d}{du}(u^n) = nu^{n-1}$

$\frac{du}{dx} = \frac{d}{dx}(1+ax) = a$

By the chain rule, $f'(x) = \frac{df}{du} \cdot \frac{du}{dx} = nu^{n-1} \cdot a = an(1+ax)^{n-1}$.

Now, we evaluate $f'(x)$ at $x=0$:

$f'(0) = an(1+a \cdot 0)^{n-1} = an(1)^{n-1}$

Assuming $n$ is such that $(1)^{n-1}$ is defined (e.g., real $n$ for $1 \neq 0$), $(1)^{n-1} = 1$.

$f'(0) = an \cdot 1 = an$.

Thus, the value of the limit is $an$.

Alternatively, using L'Hôpital's Rule on $\lim\limits_{x \to 0} \frac{(1+ax)^n - 1}{x}$ ($\frac{0}{0}$ form):

Differentiate numerator and denominator with respect to $x$:

Numerator derivative: $\frac{d}{dx}((1+ax)^n - 1) = n(1+ax)^{n-1} \cdot a - 0 = an(1+ax)^{n-1}$.

Denominator derivative: $\frac{d}{dx}(x) = 1$.

The limit becomes $\lim\limits_{x \to 0} \frac{an(1+ax)^{n-1}}{1}$.

Substitute $x=0$: $\frac{an(1+a \cdot 0)^{n-1}}{1} = \frac{an(1)^{n-1}}{1} = an$.

The value of the limit is $an$.

The correct option is (B) $an$.

Solution:

Given: The function $f(x) = \sin x + \cos x$.

To Find: The derivative of $f(x)$, denoted as $f'(x)$ or $\frac{d}{dx}[f(x)]$.

We need to find the derivative of a sum of two functions, $\sin x$ and $\cos x$. We will use the sum rule for differentiation.

The sum rule states that for two differentiable functions $u(x)$ and $v(x)$, the derivative of their sum is the sum of their derivatives:

$\frac{d}{dx}[u(x) + v(x)] = \frac{d}{dx}[u(x)] + \frac{d}{dx}[v(x)]$

In this case, let $u(x) = \sin x$ and $v(x) = \cos x$.

Find the derivative of $u(x) = \sin x$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\sin x) = \cos x$

Find the derivative of $v(x) = \cos x$ with respect to $x$:

$\frac{dv}{dx} = \frac{d}{dx}(\cos x) = -\sin x$

Now, apply the sum rule to find the derivative of $f(x) = \sin x + \cos x$:

$f'(x) = \frac{d}{dx}(\sin x + \cos x)$

$f'(x) = \frac{d}{dx}(\sin x) + \frac{d}{dx}(\cos x)$

$f'(x) = \cos x + (-\sin x)$

$f'(x) = \cos x - \sin x$

Thus, the derivative of $f(x) = \sin x + \cos x$ is $\cos x - \sin x$.

The correct option is (B) $\cos x - \sin x$.

Solution:

We need to evaluate the limit $\lim\limits_{x \to \infty} (\sqrt{x^2+x} - x)$.

As $x \to \infty$, $\sqrt{x^2+x} \approx \sqrt{x^2} = x$. So the expression is of the indeterminate form $\infty - \infty$.

To handle this indeterminate form, we can multiply the expression by its conjugate, $\sqrt{x^2+x} + x$, divided by itself:

$\lim\limits_{x \to \infty} (\sqrt{x^2+x} - x) = \lim\limits_{x \to \infty} (\sqrt{x^2+x} - x) \cdot \frac{\sqrt{x^2+x} + x}{\sqrt{x^2+x} + x}$

Use the difference of squares formula, $(a-b)(a+b) = a^2 - b^2$, in the numerator, where $a = \sqrt{x^2+x}$ and $b = x$:

Numerator $= (\sqrt{x^2+x})^2 - x^2 = (x^2+x) - x^2 = x^2 + x - x^2 = x$.

The expression becomes:

$\lim\limits_{x \to \infty} \frac{x}{\sqrt{x^2+x} + x}$

Now, as $x \to \infty$, this is of the indeterminate form $\frac{\infty}{\infty}$. To evaluate this, we can divide both the numerator and the denominator by the highest power of $x$ in the denominator, which is $x$. Note that for $x > 0$, $\sqrt{x^2} = x$, so we can divide terms inside the square root by $x^2$.

Divide numerator and denominator by $x$:

$\lim\limits_{x \to \infty} \frac{\frac{x}{x}}{\frac{\sqrt{x^2+x}}{x} + \frac{x}{x}}$

For the term $\frac{\sqrt{x^2+x}}{x}$, since $x > 0$ as $x \to \infty$, we can write $x = \sqrt{x^2}$:

$\frac{\sqrt{x^2+x}}{x} = \frac{\sqrt{x^2+x}}{\sqrt{x^2}} = \sqrt{\frac{x^2+x}{x^2}} = \sqrt{\frac{x^2}{x^2} + \frac{x}{x^2}} = \sqrt{1 + \frac{1}{x}}$

The limit expression becomes:

$\lim\limits_{x \to \infty} \frac{1}{\sqrt{1 + \frac{1}{x}} + 1}$

Now, evaluate the limit by substituting $\lim\limits_{x \to \infty} \frac{1}{x} = 0$:

$= \frac{1}{\sqrt{1 + 0} + 1}$

$= \frac{1}{\sqrt{1} + 1}$

$= \frac{1}{1 + 1}$

$= \frac{1}{2}$

The value of the limit is $\frac{1}{2}$.

The correct option is (B) 1/2.

Solution:

A polynomial function is defined as a function of the form $P(x) = c_n x^n + c_{n-1} x^{n-1} + ... + c_1 x + c_0$, where $c_i$ are constants and $n$ is a non-negative integer.

Polynomial functions are known to be continuous for all real numbers $x$.

For any continuous function $f(x)$, the limit as $x$ approaches a point $a$ is equal to the value of the function at that point.

That is, if $f$ is continuous at $a$, then $\lim\limits_{x \to a} f(x) = f(a)$.

Since $P(x)$ is a polynomial function, it is continuous at any real number $a$.

Therefore, the limit of $P(x)$ as $x \to a$ is equal to $P(a)$.

The limit of a polynomial function $P(x)$ as $x \to a$ is equal to $P(a)$.

The correct option is (B).

Solution:

We want to evaluate the limit $\lim\limits_{x \to 0} \frac{\tan(ax)}{bx}$.

As $x \to 0$, $\tan(ax) \to \tan(a \cdot 0) = \tan(0) = 0$ and $bx \to b \cdot 0 = 0$ (since $b \neq 0$). This is an indeterminate form of type $\frac{0}{0}$.

We can use the standard trigonometric limit $\lim\limits_{\theta \to 0} \frac{\tan(\theta)}{\theta} = 1$.

We can rewrite the given expression as follows:

$\frac{\tan(ax)}{bx} = \frac{1}{b} \cdot \frac{\tan(ax)}{x}$

To match the form of the standard limit, we multiply and divide the term $\frac{\tan(ax)}{x}$ by $a$ (assuming $a \neq 0$; if $a=0$, $\tan(0)/bx = 0/bx = 0$ for $x \neq 0$, so the limit is 0. In this case, $a/b = 0/b = 0$, so the formula holds):

$\frac{\tan(ax)}{x} = \frac{\tan(ax)}{ax} \cdot a$

Now substitute this back into the limit expression:

$\lim\limits_{x \to 0} \frac{\tan(ax)}{bx} = \lim\limits_{x \to 0} \left( \frac{1}{b} \cdot \frac{\tan(ax)}{ax} \cdot a \right)$

We can pull the constant factor $\frac{a}{b}$ out of the limit:

$= \frac{a}{b} \lim\limits_{x \to 0} \frac{\tan(ax)}{ax}$

Let $\theta = ax$. As $x \to 0$, $ax \to 0$, so $\theta \to 0$. The limit becomes:

$= \frac{a}{b} \lim\limits_{\theta \to 0} \frac{\tan(\theta)}{\theta}$

Using the standard limit $\lim\limits_{\theta \to 0} \frac{\tan(\theta)}{\theta} = 1$, we get:

$= \frac{a}{b} \cdot 1 = \frac{a}{b}$

The value of the limit is $\frac{a}{b}$.

The correct option is (A).

Solution:

We are asked to find the derivative of $f(x) = \sin^2 x$.

We can write $f(x)$ as $f(x) = (\sin x)^2$.

This is a composite function. Let $u = \sin x$. Then $f(x) = u^2$.

We use the Chain Rule, which states that if $y = g(u)$ and $u = h(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

Here, the outer function is $g(u) = u^2$, and the inner function is $h(x) = \sin x$.

First, find the derivative of the outer function with respect to $u$:

$\frac{dy}{du} = \frac{d}{du}(u^2) = 2u$

Next, find the derivative of the inner function with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\sin x) = \cos x$

Now, apply the Chain Rule:

$f'(x) = \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$f'(x) = (2u) \cdot (\cos x)$

Substitute $u = \sin x$ back into the expression:

$f'(x) = 2(\sin x)(\cos x)$

Using the double angle trigonometric identity $\sin(2x) = 2 \sin x \cos x$, we can also write the derivative as:

$f'(x) = \sin(2x)$

The derivative of $f(x) = \sin^2 x$ is $2 \sin x \cos x$, which is equal to $\sin 2x$.

The correct option is (C).

Solution:

The assertion states that if a function $f(x)$ is differentiable at a point $a$, then it is continuous at $a$.

The reason states that differentiability implies continuity.

A function $f(x)$ is differentiable at a point $a$ if the limit $\lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$ exists.

A function $f(x)$ is continuous at a point $a$ if $\lim\limits_{x \to a} f(x) = f(a)$. This is equivalent to $\lim\limits_{h \to 0} f(a+h) = f(a)$, or $\lim\limits_{h \to 0} (f(a+h) - f(a)) = 0$.

Consider the expression $f(a+h) - f(a)$ for $h \neq 0$:

$f(a+h) - f(a) = \frac{f(a+h) - f(a)}{h} \cdot h$

Now, take the limit as $h \to 0$ on both sides:

$\lim\limits_{h \to 0} (f(a+h) - f(a)) = \lim\limits_{h \to 0} \left( \frac{f(a+h) - f(a)}{h} \cdot h \right)$

Using the property that the limit of a product is the product of the limits (provided the individual limits exist):

$\lim\limits_{h \to 0} (f(a+h) - f(a)) = \left( \lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h} \right) \cdot \left( \lim\limits_{h \to 0} h \right)$

Since $f$ is differentiable at $a$, $\lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h} = f'(a)$, which exists and is a finite number.

Also, $\lim\limits_{h \to 0} h = 0$.

So, $\lim\limits_{h \to 0} (f(a+h) - f(a)) = f'(a) \cdot 0 = 0$.

This shows that $\lim\limits_{h \to 0} f(a+h) = f(a)$, which is the condition for continuity at $a$.

Therefore, if a function is differentiable at a point, it must be continuous at that point.

Assertion (A): If a function $f(x)$ is differentiable at a point $a$, then it is continuous at $a$. This statement is true.

Reason (R): Differentiability implies continuity. This statement is also true, and it is the general principle that proves the assertion.

The reason directly explains why the assertion is true.

Both A and R are true and R is the correct explanation of A.

The correct option is (A).

Solution:

We want to find the derivative of $f(x) = \log_{10} x$.

The derivative of a logarithmic function with base $b$ is given by the formula:

$\frac{d}{dx} (\log_b x) = \frac{1}{x \ln b}$

In this case, the base is $b=10$.

So, applying the formula, we get:

$\frac{d}{dx} (\log_{10} x) = \frac{1}{x \ln 10}$

Recall that $\ln 10$ is the natural logarithm of 10, which can also be written as $\log_e 10$.

Therefore, the derivative is $\frac{1}{x \log_e 10}$.

The derivative of $f(x) = \log_{10} x$ is $\frac{1}{x \log_e 10}$.

The correct option is (B).

Solution:

We need to evaluate the sum of two limits: $\lim\limits_{x \to 0} \frac{\sin x}{x}$ and $\lim\limits_{x \to 1} \frac{x^2-1}{x-1}$.

Let's evaluate the first limit:

$\lim\limits_{x \to 0} \frac{\sin x}{x}$

This is a standard trigonometric limit, and its value is known to be 1.

So, $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$.

Now, let's evaluate the second limit:

$\lim\limits_{x \to 1} \frac{x^2-1}{x-1}$

As $x \to 1$, the numerator approaches $1^2 - 1 = 0$ and the denominator approaches $1 - 1 = 0$. This is an indeterminate form $\frac{0}{0}$.

We can factor the numerator $x^2 - 1$ as a difference of squares: $x^2 - 1 = (x-1)(x+1)$.

So, for $x \neq 1$, the expression $\frac{x^2-1}{x-1}$ can be simplified:

$\frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} = x+1$

Now, we can evaluate the limit of the simplified expression:

$\lim\limits_{x \to 1} (x+1)$

Since $x+1$ is a polynomial function, its limit as $x \to 1$ is found by direct substitution:

$\lim\limits_{x \to 1} (x+1) = 1 + 1 = 2$.

So, $\lim\limits_{x \to 1} \frac{x^2-1}{x-1} = 2$.

Finally, we add the values of the two limits:

$\lim\limits_{x \to 0} \frac{\sin x}{x} + \lim\limits_{x \to 1} \frac{x^2-1}{x-1} = 1 + 2 = 3$.

The value of the expression is 3.

The correct option is (A).

Solution:

We are asked to find the derivative of $f(x) = x^3 \cos x$.

This function is a product of two functions: $u(x) = x^3$ and $v(x) = \cos x$.

We need to use the Product Rule for differentiation, which states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.

First, find the derivative of $u(x)$: $u'(x) = \frac{d}{dx}(x^3)$.

Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$, we get:

$u'(x) = 3x^{3-1} = 3x^2$.

Next, find the derivative of $v(x)$: $v'(x) = \frac{d}{dx}(\cos x)$.

The derivative of $\cos x$ is $-\sin x$.

$v'(x) = -\sin x$.

Now, apply the Product Rule:

$f'(x) = u'(x)v(x) + u(x)v'(x)$

$f'(x) = (3x^2)(\cos x) + (x^3)(-\sin x)$

$f'(x) = 3x^2 \cos x - x^3 \sin x$.

The derivative of $f(x) = x^3 \cos x$ is $3x^2 \cos x - x^3 \sin x$.

The correct option is (A).

Solution:

We want to evaluate the limit $\lim\limits_{x \to \infty} \frac{ax+b}{cx+d}$.

As $x \to \infty$, both the numerator ($ax+b$) and the denominator ($cx+d$) tend towards infinity (assuming $a \neq 0$ and $c \neq 0$). This is an indeterminate form of type $\frac{\infty}{\infty}$.

To evaluate limits of rational functions as $x \to \infty$, we can divide both the numerator and the denominator by the highest power of $x$ present in the denominator. In this case, the highest power of $x$ in the denominator $cx+d$ is $x^1$ (since $c \neq 0$).

Divide the numerator and the denominator by $x$:

$\frac{ax+b}{cx+d} = \frac{\frac{ax+b}{x}}{\frac{cx+d}{x}}$

$= \frac{\frac{ax}{x} + \frac{b}{x}}{\frac{cx}{x} + \frac{d}{x}}$

$= \frac{a + \frac{b}{x}}{c + \frac{d}{x}}$

Now, we evaluate the limit of this simplified expression as $x \to \infty$:

$\lim\limits_{x \to \infty} \frac{ax+b}{cx+d} = \lim\limits_{x \to \infty} \frac{a + \frac{b}{x}}{c + \frac{d}{x}}$

Using the property that $\lim\limits_{x \to \infty} \frac{k}{x^p} = 0$ for any constant $k$ and $p > 0$, we have:

$\lim\limits_{x \to \infty} \frac{b}{x} = 0$ and $\lim\limits_{x \to \infty} \frac{d}{x} = 0$.

Also, the limit of a constant is the constant itself: $\lim\limits_{x \to \infty} a = a$ and $\lim\limits_{x \to \infty} c = c$.

So, the limit becomes:

$\lim\limits_{x \to \infty} \frac{a + \frac{b}{x}}{c + \frac{d}{x}} = \frac{\lim\limits_{x \to \infty} (a + \frac{b}{x})}{\lim\limits_{x \to \infty} (c + \frac{d}{x})}$

$= \frac{\lim\limits_{x \to \infty} a + \lim\limits_{x \to \infty} \frac{b}{x}}{\lim\limits_{x \to \infty} c + \lim\limits_{x \to \infty} \frac{d}{x}}$

$= \frac{a + 0}{c + 0}$

$= \frac{a}{c}$

Note: If $a=0$, the limit is $\lim\limits_{x \to \infty} \frac{b}{cx+d}$. Since $c \neq 0$, as $x \to \infty$, the denominator approaches $\pm \infty$, so the limit is 0. Our formula $\frac{a}{c} = \frac{0}{c} = 0$ still holds.

The value of the limit is $\frac{a}{c}$.

The correct option is (A).

Solution:

We need to check if the given limits are of the indeterminate form $\frac{0}{0}$ by substituting the limiting value of $x$ into the numerator and the denominator.

For option (A): $\lim\limits_{x \to 0} \frac{x^2}{x}$

Numerator as $x \to 0$: $x^2 \to 0^2 = 0$.

Denominator as $x \to 0$: $x \to 0$.

The form is $\frac{0}{0}$.

For option (B): $\lim\limits_{x \to 1} \frac{x-1}{x^2-1}$

Numerator as $x \to 1$: $x-1 \to 1-1 = 0$.

Denominator as $x \to 1$: $x^2-1 \to 1^2-1 = 1-1 = 0$.

The form is $\frac{0}{0}$.

For option (C): $\lim\limits_{x \to 0} \frac{\sin x}{x}$

Numerator as $x \to 0$: $\sin x \to \sin(0) = 0$.

Denominator as $x \to 0$: $x \to 0$.

The form is $\frac{0}{0}$.

Since all three limits result in the indeterminate form $\frac{0}{0}$ when the limit is directly evaluated, all of the given options are correct examples of the $0/0$ indeterminate form.

All of the given limits are of the indeterminate form $0/0$.

The correct option is (D).

Solution:

We need to find the derivative of $f(x) = \log_a x$.

We can use the change of base formula for logarithms to convert the logarithm with base $a$ to the natural logarithm (base $e$). The formula is $\log_b x = \frac{\log_c x}{\log_c b}$. Using base $c=e$, we have:

$\log_a x = \frac{\log_e x}{\log_e a} = \frac{\ln x}{\ln a}$

So, $f(x) = \frac{\ln x}{\ln a}$. Since $a$ is a constant base, $\ln a$ is a constant.

Now, we can find the derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx} \left( \frac{\ln x}{\ln a} \right)$

Since $\frac{1}{\ln a}$ is a constant, we can take it out of the differentiation:

$f'(x) = \frac{1}{\ln a} \frac{d}{dx} (\ln x)$

The derivative of $\ln x$ with respect to $x$ is $\frac{1}{x}$.

Substituting this into the expression, we get:

$f'(x) = \frac{1}{\ln a} \cdot \frac{1}{x} = \frac{1}{x \ln a}$

Since $\ln a$ is the same as $\log_e a$, we can write the derivative as $\frac{1}{x \log_e a}$.

The derivative of $f(x) = \log_a x$ is $\frac{1}{x \log_e a}$.

The correct option is (B).

Solution:

We want to evaluate the limit $\lim\limits_{x \to 0} \frac{1 - \cos x}{x \sin x}$.

As $x \to 0$, the numerator $1 - \cos x \to 1 - \cos(0) = 1 - 1 = 0$.

As $x \to 0$, the denominator $x \sin x \to 0 \cdot \sin(0) = 0 \cdot 0 = 0$.

This is an indeterminate form of type $\frac{0}{0}$.

We can rewrite the expression using standard trigonometric limits.

Recall the standard limits:

$\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$

$\lim\limits_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$

Let's manipulate the given expression to involve these standard forms:

$\frac{1 - \cos x}{x \sin x} = \frac{1 - \cos x}{x \cdot x \cdot \frac{\sin x}{x}}$

We can rearrange the terms:

$\frac{1 - \cos x}{x \sin x} = \frac{1 - \cos x}{x^2} \cdot \frac{x}{\sin x}$

Taking the limit as $x \to 0$:

$\lim\limits_{x \to 0} \frac{1 - \cos x}{x \sin x} = \lim\limits_{x \to 0} \left( \frac{1 - \cos x}{x^2} \cdot \frac{x}{\sin x} \right)$

Using the property that the limit of a product is the product of the limits (if they exist):

$= \left( \lim\limits_{x \to 0} \frac{1 - \cos x}{x^2} \right) \cdot \left( \lim\limits_{x \to 0} \frac{x}{\sin x} \right)$

We know $\lim\limits_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$.

For the second limit, $\lim\limits_{x \to 0} \frac{x}{\sin x}$, we can use the reciprocal of the standard limit: $\lim\limits_{x \to 0} \frac{x}{\sin x} = \frac{1}{\lim\limits_{x \to 0} \frac{\sin x}{x}} = \frac{1}{1} = 1$.

Substitute these values back into the expression:

$= \left( \frac{1}{2} \right) \cdot (1) = \frac{1}{2}$

The value of the limit is $\frac{1}{2}$.

The correct option is (A).

Solution:

We are asked to find the derivative of $f(x) = \frac{\cos x}{x}$.

This function is a quotient of two functions: $u(x) = \cos x$ and $v(x) = x$.

We need to use the Quotient Rule for differentiation, which states that if $f(x) = \frac{u(x)}{v(x)}$, then the derivative $f'(x)$ is given by:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

First, find the derivative of the numerator $u(x) = \cos x$:

$u'(x) = \frac{d}{dx}(\cos x) = -\sin x$

Next, find the derivative of the denominator $v(x) = x$:

$v'(x) = \frac{d}{dx}(x) = 1$

Now, apply the Quotient Rule formula:

$f'(x) = \frac{(-\sin x)(x) - (\cos x)(1)}{x^2}$

$f'(x) = \frac{-x \sin x - \cos x}{x^2}$

The derivative of $f(x) = \frac{\cos x}{x}$ is $\frac{-x \sin x - \cos x}{x^2}$.

The correct option is (B).

Solution:

We are asked to evaluate the limit $\lim\limits_{h \to 0} \frac{(x+h)^n - x^n}{h}$.

This limit expression is the fundamental definition of the derivative of a function $f(x)$ with respect to $x$.

Specifically, the derivative of a function $f(x)$ at a point $x$ is defined as:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Comparing this definition with the given limit expression, we can see that it corresponds to the derivative of the function $f(x) = x^n$.

We need to find the derivative of $f(x) = x^n$. Using the power rule for differentiation, which states that $\frac{d}{dx}(x^n) = nx^{n-1}$, we find the derivative:

$\frac{d}{dx}(x^n) = nx^{n-1}$

So, the value of the limit $\lim\limits_{h \to 0} \frac{(x+h)^n - x^n}{h}$ is equal to the derivative of $x^n$, which is $nx^{n-1}$.

The question also states that this result is effectively asking for the derivative of $x^n$ evaluated at $x=a$.

To evaluate the derivative $nx^{n-1}$ at $x=a$, we substitute $a$ for $x$:

$[nx^{n-1}]_{x=a} = na^{n-1}$

The value of the limit (which is the derivative of $x^n$ evaluated at $x=a$) is $na^{n-1}$.

The correct option is (A).

Solution:

We are asked to find the derivative of the function $f(x) = \sin(ax+b)$.

This is a composite function. We can apply the Chain Rule.

Let the outer function be $g(u) = \sin u$ and the inner function be $h(x) = ax+b$. Then $f(x) = g(h(x))$.

The Chain Rule states that the derivative of a composite function $f(x) = g(h(x))$ is given by $f'(x) = g'(h(x)) \cdot h'(x)$.

First, find the derivative of the outer function $g(u)$ with respect to $u$:

$g'(u) = \frac{d}{du}(\sin u) = \cos u$

Next, find the derivative of the inner function $h(x)$ with respect to $x$:

$h'(x) = \frac{d}{dx}(ax+b)$

Using the rules for differentiation of linear terms and constants:

$h'(x) = \frac{d}{dx}(ax) + \frac{d}{dx}(b) = a \frac{d}{dx}(x) + 0 = a(1) = a$

Now, apply the Chain Rule $f'(x) = g'(h(x)) \cdot h'(x)$:

$f'(x) = \cos(ax+b) \cdot a$

Rearranging the terms, we get:

$f'(x) = a \cos(ax+b)$

The derivative of $f(x) = \sin(ax+b)$ is $a \cos(ax+b)$.

The correct option is (B).

Solution:

We want to evaluate the limit $\lim\limits_{x \to 0} \frac{\sin^2 x}{x}$.

As $x \to 0$, the numerator $\sin^2 x$ approaches $\sin^2(0) = 0^2 = 0$.

As $x \to 0$, the denominator $x$ approaches $0$.

This is an indeterminate form of type $\frac{0}{0}$.

We can rewrite the expression as $\frac{\sin x \cdot \sin x}{x}$.

We can group the terms to use the standard limit $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$.

$\frac{\sin^2 x}{x} = \frac{\sin x}{x} \cdot \sin x$

Now, take the limit as $x \to 0$:

$\lim\limits_{x \to 0} \frac{\sin^2 x}{x} = \lim\limits_{x \to 0} \left( \frac{\sin x}{x} \cdot \sin x \right)$

Using the property that the limit of a product is the product of the limits (if they exist separately):

$= \left( \lim\limits_{x \to 0} \frac{\sin x}{x} \right) \cdot \left( \lim\limits_{x \to 0} \sin x \right)$

We know that $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$.

Also, the sine function is continuous at $x=0$, so $\lim\limits_{x \to 0} \sin x = \sin(0) = 0$.

Substitute these values back into the expression:

$= 1 \cdot 0 = 0$

The value of the limit is $0$.

The correct option is (A).

Solution:

We are asked to find the derivative of $f(x) = \text{cosec } x$.

The derivative of the cosecant function is a standard differentiation formula.

The derivative of $\text{cosec } x$ with respect to $x$ is given by:

$\frac{d}{dx}(\text{cosec } x) = -\text{cosec } x \cot x$

The derivative of $\text{cosec } x$ is $-\text{cosec } x \cot x$.

The correct option is (B).

Solution:

We want to evaluate the limit $\lim\limits_{x \to 2} \frac{\sqrt{x+2} - 2}{x - 2}$.

As $x \to 2$, the numerator $\sqrt{x+2} - 2 \to \sqrt{2+2} - 2 = \sqrt{4} - 2 = 2 - 2 = 0$.

As $x \to 2$, the denominator $x - 2 \to 2 - 2 = 0$.

This is an indeterminate form of type $\frac{0}{0}$.

To evaluate this limit, we can rationalize the numerator by multiplying both the numerator and the denominator by the conjugate of the numerator, which is $\sqrt{x+2} + 2$.

$\lim\limits_{x \to 2} \frac{\sqrt{x+2} - 2}{x - 2} = \lim\limits_{x \to 2} \frac{\sqrt{x+2} - 2}{x - 2} \cdot \frac{\sqrt{x+2} + 2}{\sqrt{x+2} + 2}$

In the numerator, we use the difference of squares formula $(a-b)(a+b) = a^2 - b^2$ where $a = \sqrt{x+2}$ and $b = 2$.

$= \lim\limits_{x \to 2} \frac{(\sqrt{x+2})^2 - 2^2}{(x - 2)(\sqrt{x+2} + 2)}$

$= \lim\limits_{x \to 2} \frac{(x+2) - 4}{(x - 2)(\sqrt{x+2} + 2)}$

$= \lim\limits_{x \to 2} \frac{x - 2}{(x - 2)(\sqrt{x+2} + 2)}$

For $x \neq 2$, we can cancel the common factor $(x-2)$ in the numerator and the denominator.

$= \lim\limits_{x \to 2} \frac{1}{\sqrt{x+2} + 2}$

Now, substitute $x = 2$ into the simplified expression (since the denominator is no longer zero at $x=2$).

$= \frac{1}{\sqrt{2+2} + 2}$

$= \frac{1}{\sqrt{4} + 2}$

$= \frac{1}{2 + 2}$

$= \frac{1}{4}$

The value of the limit is $\frac{1}{4}$.

The correct option is (B).

Solution:

We need to find the derivative of the function $f(x) = \tan(ax)$.

This is a composite function, so we will use the Chain Rule.

Let $g(u) = \tan u$ be the outer function and $h(x) = ax$ be the inner function. Then $f(x) = g(h(x))$.

The Chain Rule states that the derivative of $f(x)$ is $f'(x) = g'(h(x)) \cdot h'(x)$.

First, find the derivative of the outer function $g(u)$ with respect to $u$:

$g'(u) = \frac{d}{du}(\tan u) = \sec^2 u$

Next, find the derivative of the inner function $h(x)$ with respect to $x$:

$h'(x) = \frac{d}{dx}(ax)$

Using the rule for the derivative of $kx$, we have:

$h'(x) = a \frac{d}{dx}(x) = a \cdot 1 = a$

Now, apply the Chain Rule $f'(x) = g'(h(x)) \cdot h'(x)$:

$f'(x) = \sec^2(ax) \cdot a$

Rearranging the terms, we get:

$f'(x) = a \sec^2(ax)$

The derivative of $f(x) = \tan(ax)$ is $a \sec^2(ax)$.

The correct option is (A).

Solution:

We need to identify which of the given options represents a standard limit involving a trigonometric function.

Let's examine each option:

(A) $\lim\limits_{x \to 0} (1+x)^{1/x}$

This is a standard limit related to the definition of the mathematical constant $e$. Its value is $\lim\limits_{x \to 0} (1+x)^{1/x} = e$. This is not a trigonometric limit.

(B) $\lim\limits_{x \to 0} \frac{e^x - 1}{x}$

This is a standard limit involving the exponential function $e^x$. Its value is $\lim\limits_{x \to 0} \frac{e^x - 1}{x} = 1$. This is not a trigonometric limit.

(C) $\lim\limits_{x \to 0} \frac{\sin x}{x}$

This is a fundamental standard limit involving the trigonometric function $\sin x$. Its value is $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$. This is a standard trigonometric limit.

(D) $\lim\limits_{x \to a} \frac{x^n - a^n}{x - a}$

This limit represents the definition of the derivative of the function $f(x) = x^n$ at $x=a$. Its value is $\lim\limits_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}$. This is related to power functions and derivatives, not primarily a trigonometric limit.

Based on the analysis, only option (C) is a standard limit of a trigonometric function.

The correct option is (C).

Solution:

We need to find the derivative of the function $f(x) = x^5 + 3$.

We can use the properties of differentiation:

The derivative of a sum of functions is the sum of their derivatives: $\frac{d}{dx}[g(x) + h(x)] = g'(x) + h'(x)$.

The derivative of a power function $x^n$ is $nx^{n-1}$: $\frac{d}{dx}(x^n) = nx^{n-1}$.

The derivative of a constant $c$ is $0$: $\frac{d}{dx}(c) = 0$.

Applying these rules to $f(x) = x^5 + 3$:

$f'(x) = \frac{d}{dx}(x^5 + 3)$

$f'(x) = \frac{d}{dx}(x^5) + \frac{d}{dx}(3)$

Using the power rule for $x^5$ (with $n=5$):

$\frac{d}{dx}(x^5) = 5x^{5-1} = 5x^4$

Using the rule for the derivative of a constant for 3:

$\frac{d}{dx}(3) = 0$

Combining these results:

$f'(x) = 5x^4 + 0$

$f'(x) = 5x^4$

The derivative of $f(x) = x^5 + 3$ is $5x^4$.

The correct option is (A).

Solution:

We want to evaluate the limit $\lim\limits_{x \to 0} \frac{\sin 4x}{2x}$.

As $x \to 0$, the numerator $\sin 4x \to \sin(4 \cdot 0) = \sin(0) = 0$.

As $x \to 0$, the denominator $2x \to 2 \cdot 0 = 0$.

This is an indeterminate form of type $\frac{0}{0}$.

We can use the standard trigonometric limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

To use this standard limit, the argument of the sine function must be the same as the denominator.

In our case, the argument of the sine function is $4x$, but the denominator is $2x$.

We can rewrite the expression by multiplying and dividing by 2:

$\frac{\sin 4x}{2x} = \frac{\sin 4x}{2x} \cdot \frac{2}{2} = \frac{\sin 4x}{4x} \cdot 2$

Now, take the limit as $x \to 0$:

$\lim\limits_{x \to 0} \frac{\sin 4x}{2x} = \lim\limits_{x \to 0} \left( \frac{\sin 4x}{4x} \cdot 2 \right)$

Using the properties of limits, we can factor out the constant 2:

$= 2 \lim\limits_{x \to 0} \frac{\sin 4x}{4x}$

Let $\theta = 4x$. As $x \to 0$, $4x \to 0$, so $\theta \to 0$. The limit becomes:

$= 2 \lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta}$

Using the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

$= 2 \cdot 1 = 2$

The value of the limit is $2$.

The correct option is (B).

Solution:

We need to find the derivative of the function $f(x) = \sec(3x)$.

This is a composite function, so we will use the Chain Rule.

Let the outer function be $g(u) = \sec u$ and the inner function be $h(x) = 3x$. Then $f(x) = g(h(x))$.

The Chain Rule states that the derivative of a composite function $f(x) = g(h(x))$ is given by $f'(x) = g'(h(x)) \cdot h'(x)$.

First, find the derivative of the outer function $g(u)$ with respect to $u$:

$g'(u) = \frac{d}{du}(\sec u) = \sec u \tan u$

Next, find the derivative of the inner function $h(x)$ with respect to $x$:

$h'(x) = \frac{d}{dx}(3x)$

Using the rule for the derivative of $kx$, we have:

$h'(x) = 3 \frac{d}{dx}(x) = 3 \cdot 1 = 3$

Now, apply the Chain Rule $f'(x) = g'(h(x)) \cdot h'(x)$:

$f'(x) = \sec(3x) \tan(3x) \cdot 3$

Rearranging the terms, we get:

$f'(x) = 3 \sec(3x) \tan(3x)$

The derivative of $f(x) = \sec(3x)$ is $3 \sec(3x) \tan(3x)$.

The correct option is (B).

Solution:

We want to evaluate the limit $\lim\limits_{x \to 0} \frac{1 - \cos x}{x}$.

As $x \to 0$, the numerator $1 - \cos x \to 1 - \cos(0) = 1 - 1 = 0$.

As $x \to 0$, the denominator $x \to 0$.

This is an indeterminate form of type $\frac{0}{0}$.

We can use the standard trigonometric limit $\lim\limits_{x \to 0} \frac{1 - \cos x}{x} = 0$.

Alternatively, we can multiply the numerator and denominator by the conjugate of the numerator, $1 + \cos x$:

$\lim\limits_{x \to 0} \frac{1 - \cos x}{x} = \lim\limits_{x \to 0} \frac{1 - \cos x}{x} \cdot \frac{1 + \cos x}{1 + \cos x}$

$= \lim\limits_{x \to 0} \frac{1 - \cos^2 x}{x(1 + \cos x)}$

Using the identity $\sin^2 x + \cos^2 x = 1$, so $1 - \cos^2 x = \sin^2 x$:

$= \lim\limits_{x \to 0} \frac{\sin^2 x}{x(1 + \cos x)}$

We can rewrite this as:

$= \lim\limits_{x \to 0} \left( \frac{\sin x}{x} \cdot \frac{\sin x}{1 + \cos x} \right)$

Using the property that the limit of a product is the product of the limits:

$= \left( \lim\limits_{x \to 0} \frac{\sin x}{x} \right) \cdot \left( \lim\limits_{x \to 0} \frac{\sin x}{1 + \cos x} \right)$

We know the standard limit $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$.

For the second part, substitute $x=0$ (since the denominator is non-zero at $x=0$):

$\lim\limits_{x \to 0} \frac{\sin x}{1 + \cos x} = \frac{\sin(0)}{1 + \cos(0)} = \frac{0}{1 + 1} = \frac{0}{2} = 0$.

Multiplying the results:

$= 1 \cdot 0 = 0$

The value of the limit is $0$.

The correct option is (A).

Solution:

We are asked to find the derivative of $f(x) = \frac{\sin x}{\cos x}$.

We know that $\frac{\sin x}{\cos x} = \tan x$.

So, we need to find the derivative of $f(x) = \tan x$.

The derivative of the tangent function is a standard differentiation formula:

$\frac{d}{dx}(\tan x) = \sec^2 x$

Alternatively, we can use the Quotient Rule for differentiation.

Let $u(x) = \sin x$ and $v(x) = \cos x$.

Then $u'(x) = \frac{d}{dx}(\sin x) = \cos x$ and $v'(x) = \frac{d}{dx}(\cos x) = -\sin x$.

Using the Quotient Rule: $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

$f'(x) = \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{(\cos x)^2}$

$f'(x) = \frac{\cos^2 x + \sin^2 x}{\cos^2 x}$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$:

$f'(x) = \frac{1}{\cos^2 x}$

Since $\sec x = \frac{1}{\cos x}$, we have $\sec^2 x = \frac{1}{\cos^2 x}$.

Thus, $f'(x) = \sec^2 x$.

The derivative of $f(x) = \frac{\sin x}{\cos x}$ is $\sec^2 x$.

The correct option is (B).

Solution:

We are given the function $R(T) = \frac{T^2 - 4}{T - 2}$ for $T \neq 2$ and asked to evaluate the limit of $R(T)$ as $T \to 2$.

We need to evaluate $\lim\limits_{T \to 2} \frac{T^2 - 4}{T - 2}$.

As $T \to 2$, the numerator $T^2 - 4 \to 2^2 - 4 = 4 - 4 = 0$.

As $T \to 2$, the denominator $T - 2 \to 2 - 2 = 0$.

This is an indeterminate form of type $\frac{0}{0}$.

To evaluate the limit, we can factor the numerator, which is a difference of squares:

$T^2 - 4 = T^2 - 2^2 = (T - 2)(T + 2)$.

So, for $T \neq 2$, the function can be simplified:

$R(T) = \frac{(T - 2)(T + 2)}{T - 2}$

Now, we can evaluate the limit of the simplified expression. Since we are considering the limit as $T \to 2$, we are looking at values of $T$ close to 2 but not equal to 2. Therefore, $T - 2 \neq 0$, and we can cancel the $(T-2)$ term in the numerator and denominator.

$\lim\limits_{T \to 2} \frac{(T - 2)(T + 2)}{T - 2} = \lim\limits_{T \to 2} (T + 2)$

The expression is now a polynomial $T+2$, which is continuous everywhere. The limit of a polynomial as the variable approaches a value is equal to the value of the polynomial at that point (direct substitution).

$\lim\limits_{T \to 2} (T + 2) = 2 + 2 = 4$

Let's examine the options:

(A) $\lim\limits_{T \to 2} (T+2)$: This is the limit of the simplified expression, which is a correct step.

(B) $\lim\limits_{T \to 2} \frac{(T-2)(T+2)}{T-2}$: This shows the factored form of the expression inside the limit, which is also a correct step in the process.

(C) 4: This is the final calculated value of the limit, which is correct.

(D) All of the above steps and the result are correct: Since options (A), (B), and (C) accurately reflect correct steps or the correct result in evaluating the limit, this option is correct.

The correct option is (D).

Solution:

From the previous question (Question 95), we found that the function for the reaction rate is $R(T) = \frac{T^2 - 4}{T - 2}$ for $T \neq 2$.

We also simplified this expression for $T \neq 2$ as:

$R(T) = \frac{(T - 2)(T + 2)}{T - 2} = T + 2$

So, for $T \neq 2$, $R(T)$ behaves like the function $T+2$.

The instantaneous rate of change of $R(T)$ with respect to $T$ is given by the derivative $R'(T)$. Since $R(T) = T+2$ for $T \neq 2$, we find the derivative of $T+2$ with respect to $T$.

The derivative of a sum is the sum of the derivatives:

$R'(T) = \frac{d}{dT}(T + 2)$

$R'(T) = \frac{d}{dT}(T) + \frac{d}{dT}(2)$

The derivative of $T$ with respect to $T$ is 1.

The derivative of a constant (2) with respect to $T$ is 0.

$R'(T) = 1 + 0$

$R'(T) = 1$

This derivative is valid for $T \neq 2$.

Let's examine the given options in light of our calculation:

(A) $\frac{d}{dT} (T+2)$: This is the correct expression for the derivative of the simplified function $R(T)$ for $T \neq 2$. This represents a correct step in the process.

(B) 1: This is the correct value of the derivative $R'(T)$ for $T \neq 2$. This represents the correct result.

(C) 0: This is incorrect.

(D) All of the above steps and the result are correct: Option (A) represents a correct step (taking the derivative of the simplified function), and option (B) represents the correct result of that differentiation. Therefore, this option asserts that both the relevant step shown and the final result are correct, which aligns with our findings.

The correct option is (D).

Solution:

From the previous questions, we have the function for the reaction rate for $T \neq 2$ as:

$R(T) = \frac{T^2 - 4}{T - 2}$

For $T \neq 2$, this function simplifies to:

$R(T) = T + 2$

The instantaneous rate of change of $R(T)$ with respect to $T$ is given by the derivative, $R'(T)$.

From Question 96, we found the derivative of $R(T)$ for $T \neq 2$:

$R'(T) = \frac{d}{dT}(T + 2) = 1$

We are asked to find the instantaneous rate of change of the reaction rate at $T=3$. This means we need to evaluate $R'(T)$ at $T=3$.

Since $T=3$ is not equal to 2, the derivative $R'(T) = 1$ is applicable at this point.

Substituting $T=3$ into the expression for $R'(T)$:

$R'(3) = 1$

The instantaneous rate of change of the reaction rate at $T=3$ is 1.

The correct option is (A).

Solution:

We want to evaluate the limit $\lim\limits_{x \to \infty} \frac{\sin x}{x}$.

As $x \to \infty$, the value of $\sin x$ oscillates between $-1$ and $1$. It does not approach a specific value, but it is a bounded function.

The denominator $x$ approaches $\infty$ as $x \to \infty$.

We can use the Squeeze Theorem to evaluate this limit.

We know that for any real number $x$, the value of $\sin x$ satisfies the inequality:

$-1 \leq \sin x \leq 1$

For $x > 0$, we can divide all parts of the inequality by $x$ without changing the direction of the inequalities:

$\frac{-1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x}$

Now, let's evaluate the limits of the bounding functions as $x \to \infty$:

$\lim\limits_{x \to \infty} \frac{-1}{x} = 0$

$\lim\limits_{x \to \infty} \frac{1}{x} = 0$

Since $\lim\limits_{x \to \infty} \frac{-1}{x} = 0$ and $\lim\limits_{x \to \infty} \frac{1}{x} = 0$, and $\frac{-1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x}$ for $x > 0$, by the Squeeze Theorem (also known as the Sandwich Theorem), the limit of the middle function must also be 0.

Therefore, $\lim\limits_{x \to \infty} \frac{\sin x}{x} = 0$.

The value of the limit is 0.

The correct option is (A).

Solution:

We need to find the derivative of the function $f(x) = \sec^2 x$.

We can write $f(x)$ as $f(x) = (\sec x)^2$.

This is a composite function. We will use the Chain Rule.

Let the outer function be $g(u) = u^2$ and the inner function be $h(x) = \sec x$. Then $f(x) = g(h(x))$.

The Chain Rule states that the derivative of a composite function $f(x) = g(h(x))$ is given by $f'(x) = g'(h(x)) \cdot h'(x)$.

First, find the derivative of the outer function $g(u)$ with respect to $u$:

$g'(u) = \frac{d}{du}(u^2) = 2u$

Next, find the derivative of the inner function $h(x)$ with respect to $x$:

$h'(x) = \frac{d}{dx}(\sec x) = \sec x \tan x$

Now, apply the Chain Rule $f'(x) = g'(h(x)) \cdot h'(x)$:

$f'(x) = 2(\sec x) \cdot (\sec x \tan x)$

Simplifying the expression:

$f'(x) = 2 \sec^2 x \tan x$

The derivative of $f(x) = \sec^2 x$ is $2 \sec^2 x \tan x$.

Comparing this result with the given options, none of the options exactly match $2 \sec^2 x \tan x$. However, option (B) is $2 \sec x \tan x$. Based on the structure of similar derivative problems and common textbook exercises, it is likely that option (B) was intended to be $2 \sec^2 x \tan x$. Assuming a potential typo in the options provided, option (B) represents the closest form related to the correct derivative.

Assuming the intended answer is among the options, and based on the calculated derivative:

The correct option is (B).

Solution:

We want to evaluate the limit $\lim\limits_{x \to 0} \frac{\cos x - 1}{x}$.

As $x \to 0$, the numerator $\cos x - 1 \to \cos(0) - 1 = 1 - 1 = 0$.

As $x \to 0$, the denominator $x \to 0$.

This is an indeterminate form of type $\frac{0}{0}$.

We can rewrite the expression slightly:

$\frac{\cos x - 1}{x} = - \frac{1 - \cos x}{x}$

Now, evaluate the limit:

$\lim\limits_{x \to 0} \frac{\cos x - 1}{x} = \lim\limits_{x \to 0} \left( - \frac{1 - \cos x}{x} \right)$

Using the property that the limit of a constant times a function is the constant times the limit of the function:

$= - \lim\limits_{x \to 0} \frac{1 - \cos x}{x}$

This is a standard trigonometric limit, and its value is 0.

$\lim\limits_{x \to 0} \frac{1 - \cos x}{x} = 0$

Substituting this value back into the expression:

$= -(0) = 0$

Alternatively, we can use L'Hopital's Rule since the limit is of the form $\frac{0}{0}$.

Let $f(x) = \cos x - 1$ and $g(x) = x$.

$f'(x) = \frac{d}{dx}(\cos x - 1) = -\sin x$

$g'(x) = \frac{d}{dx}(x) = 1$

Applying L'Hopital's Rule:

$\lim\limits_{x \to 0} \frac{\cos x - 1}{x} = \lim\limits_{x \to 0} \frac{-\sin x}{1}$

Substitute $x=0$ into the resulting expression:

$= \frac{-\sin(0)}{1} = \frac{-0}{1} = 0$

The value of the limit is $0$.

The correct option is (A).

Solution:

We are asked to find the derivative of $f(x) = \sqrt{x}$.

We can rewrite $\sqrt{x}$ using exponents: $f(x) = x^{1/2}$.

We use the power rule for differentiation, which states that for any real number $n$, the derivative of $x^n$ is $\frac{d}{dx}(x^n) = nx^{n-1}$.

In this case, $n = \frac{1}{2}$.

Applying the power rule:

$f'(x) = \frac{d}{dx}(x^{1/2})$

$f'(x) = \frac{1}{2} x^{\frac{1}{2} - 1}$

$f'(x) = \frac{1}{2} x^{-\frac{1}{2}}$

We can rewrite $x^{-\frac{1}{2}}$ with a positive exponent and in radical form:

$x^{-\frac{1}{2}} = \frac{1}{x^{1/2}} = \frac{1}{\sqrt{x}}$

So, substitute this back into the derivative expression:

$f'(x) = \frac{1}{2} \cdot \frac{1}{\sqrt{x}}$

$f'(x) = \frac{1}{2\sqrt{x}}$

The derivative of $f(x) = \sqrt{x}$ is $\frac{1}{2\sqrt{x}}$.

The correct option is (A).

Solution:

The question asks us to identify which of the given functions, when differentiated, does NOT result in $\frac{1}{x}$.

Let's find the derivative of each function provided in the options:

For option (A): $f(x) = \log_e x$.

The notation $\log_e x$ is the same as $\ln x$ (natural logarithm).

The derivative of $\ln x$ is $\frac{d}{dx}(\ln x) = \frac{1}{x}$.

For option (B): $f(x) = \ln x$.

As stated above, $\ln x$ is the natural logarithm, which is $\log_e x$.

The derivative of $\ln x$ is $\frac{d}{dx}(\ln x) = \frac{1}{x}$.

For option (C): $f(x) = \log_{10} x$.

This is a logarithm with base 10. The derivative of a logarithm with base $b$ is given by the formula $\frac{d}{dx}(\log_b x) = \frac{1}{x \log_e b}$.

For $f(x) = \log_{10} x$, the base is $b=10$.

The derivative is $\frac{d}{dx}(\log_{10} x) = \frac{1}{x \log_e 10}$.

Since $\log_e 10 = \ln 10 \neq 1$, the derivative $\frac{1}{x \log_e 10}$ is not equal to $\frac{1}{x}$.

For option (D): $f(x) = \log_e x + C$, where $C$ is a constant.

Using the sum rule and the derivative of a constant:

$\frac{d}{dx}(\log_e x + C) = \frac{d}{dx}(\log_e x) + \frac{d}{dx}(C)$

$= \frac{1}{x} + 0 = \frac{1}{x}$.

We are looking for the function whose derivative is NOT $\frac{1}{x}$.

From our calculations, the derivative of $f(x) = \log_e x$ is $\frac{1}{x}$.

The derivative of $f(x) = \ln x$ is $\frac{1}{x}$.

The derivative of $f(x) = \log_{10} x$ is $\frac{1}{x \log_e 10}$, which is not $\frac{1}{x}$.

The derivative of $f(x) = \log_e x + C$ is $\frac{1}{x}$.

Therefore, the function whose derivative is NOT $\frac{1}{x}$ is $f(x) = \log_{10} x$.

The correct option is (C).

Solution:

We want to evaluate the limit $\lim\limits_{x \to 0} \frac{e^{ax} - e^{bx}}{x}$.

As $x \to 0$, the numerator $e^{ax} - e^{bx} \to e^{a \cdot 0} - e^{b \cdot 0} = e^0 - e^0 = 1 - 1 = 0$.

As $x \to 0$, the denominator $x \to 0$.

This is an indeterminate form of type $\frac{0}{0}$.

Method 1: Using Standard Limits

We can use the standard limit $\lim\limits_{y \to 0} \frac{e^y - 1}{y} = 1$.

Rewrite the expression by adding and subtracting 1 in the numerator:

$\frac{e^{ax} - e^{bx}}{x} = \frac{e^{ax} - 1 - e^{bx} + 1}{x}$

$= \frac{(e^{ax} - 1) - (e^{bx} - 1)}{x}$

$= \frac{e^{ax} - 1}{x} - \frac{e^{bx} - 1}{x}$

Now, multiply and divide each term by the constant in the exponent to match the standard limit form:

$= \frac{e^{ax} - 1}{ax} \cdot a - \frac{e^{bx} - 1}{bx} \cdot b$

Now, take the limit as $x \to 0$:

$\lim\limits_{x \to 0} \left( a \frac{e^{ax} - 1}{ax} - b \frac{e^{bx} - 1}{bx} \right)$

Using the property that the limit of a difference is the difference of the limits, and factoring out constants:

$= a \lim\limits_{x \to 0} \frac{e^{ax} - 1}{ax} - b \lim\limits_{x \to 0} \frac{e^{bx} - 1}{bx}$

Let $y = ax$. As $x \to 0$, $y \to 0$. So $\lim\limits_{x \to 0} \frac{e^{ax} - 1}{ax} = \lim\limits_{y \to 0} \frac{e^y - 1}{y} = 1$.

Let $z = bx$. As $x \to 0$, $z \to 0$. So $\lim\limits_{x \to 0} \frac{e^{bx} - 1}{bx} = \lim\limits_{z \to 0} \frac{e^z - 1}{z} = 1$.

Substituting these values:

$= a \cdot 1 - b \cdot 1 = a - b$

Method 2: Using L'Hopital's Rule

Since the limit is of the indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule. Let $f(x) = e^{ax} - e^{bx}$ and $g(x) = x$.

Find the derivatives of $f(x)$ and $g(x)$:

$f'(x) = \frac{d}{dx}(e^{ax} - e^{bx}) = a e^{ax} - b e^{bx}$

$g'(x) = \frac{d}{dx}(x) = 1$

Apply L'Hopital's Rule:

$\lim\limits_{x \to 0} \frac{e^{ax} - e^{bx}}{x} = \lim\limits_{x \to 0} \frac{a e^{ax} - b e^{bx}}{1}$

Substitute $x = 0$ into the expression:

$= \frac{a e^{a \cdot 0} - b e^{b \cdot 0}}{1} = \frac{a e^0 - b e^0}{1} = \frac{a \cdot 1 - b \cdot 1}{1} = a - b$

The value of the limit is $a-b$.

The correct option is (A).

Solution:

We need to find the derivative of the function $f(x) = \sin^3 x$.

We can rewrite $f(x)$ as $f(x) = (\sin x)^3$.

This is a composite function. We will use the Chain Rule.

Let the outer function be $g(u) = u^3$ and the inner function be $h(x) = \sin x$. Then $f(x) = g(h(x))$.

The Chain Rule states that the derivative of a composite function $f(x) = g(h(x))$ is given by $f'(x) = g'(h(x)) \cdot h'(x)$.

First, find the derivative of the outer function $g(u)$ with respect to $u$:

$g'(u) = \frac{d}{du}(u^3) = 3u^{3-1} = 3u^2$

Next, find the derivative of the inner function $h(x)$ with respect to $x$:

$h'(x) = \frac{d}{dx}(\sin x) = \cos x$

Now, apply the Chain Rule $f'(x) = g'(h(x)) \cdot h'(x)$. Substitute $u = \sin x$ into $g'(u)$:

$f'(x) = 3(\sin x)^2 \cdot (\cos x)$

$f'(x) = 3 \sin^2 x \cos x$

The derivative of $f(x) = \sin^3 x$ is $3 \sin^2 x \cos x$.

The correct option is (B).

Solution:

We are asked to find the derivative of the function $f(x) = a^x$, where $a$ is a positive constant.

The derivative of an exponential function with base $a$ is a standard result in calculus.

The formula for the derivative of $a^x$ with respect to $x$ is given by:

$\frac{d}{dx}(a^x) = a^x \ln a$

where $\ln a$ is the natural logarithm of $a$, which is $\log_e a$.

So, the derivative of $f(x) = a^x$ is $f'(x) = a^x \log_e a$.

The derivative of $f(x) = a^x$ is $a^x \log_e a$.

The correct option is (C).

Solution:

We want to evaluate the limit $\lim\limits_{x \to 0} \frac{\sin x}{x \cos x}$.

As $x \to 0$, the numerator $\sin x \to \sin(0) = 0$.

As $x \to 0$, the denominator $x \cos x \to 0 \cdot \cos(0) = 0 \cdot 1 = 0$.

This is an indeterminate form of type $\frac{0}{0}$.

We can rewrite the expression by separating the terms:

$\frac{\sin x}{x \cos x} = \frac{\sin x}{x} \cdot \frac{1}{\cos x}$

Now, we can evaluate the limit using the property that the limit of a product is the product of the limits, provided the individual limits exist:

$\lim\limits_{x \to 0} \frac{\sin x}{x \cos x} = \lim\limits_{x \to 0} \left( \frac{\sin x}{x} \cdot \frac{1}{\cos x} \right)$

$= \left( \lim\limits_{x \to 0} \frac{\sin x}{x} \right) \cdot \left( \lim\limits_{x \to 0} \frac{1}{\cos x} \right)$

We use the standard trigonometric limit:

And for the second limit, since $\cos x$ is continuous at $x=0$ and $\cos(0) \neq 0$:

Substitute these values back into the expression:

$\lim\limits_{x \to 0} \frac{\sin x}{x \cos x} = 1 \cdot 1 = 1$

The value of the limit is 1.

The correct option is (C).

We are asked to find the derivative of the function $f(x) = \frac{1}{x}$.

We can rewrite the function using negative exponents as $f(x) = x^{-1}$.

To find the derivative of a power function of the form $x^n$, we use the power rule of differentiation.

The power rule states that if $f(x) = x^n$, where $n$ is a real number, then its derivative is given by $f'(x) = nx^{n-1}$.

In our case, the function is $f(x) = x^{-1}$.

Comparing this with the standard power function $x^n$, we see that $n = -1$.

Applying the power rule, we get:

$f'(x) = n \cdot x^{n-1}$

$f'(x) = (-1) \cdot x^{(-1)-1}$

$f'(x) = -1 \cdot x^{-2}$

$f'(x) = -x^{-2}$

The derivative can also be written in terms of positive exponents:

Recall that $x^{-2} = \frac{1}{x^2}$.

So, $f'(x) = -x^{-2} = -\frac{1}{x^2}$.

Comparing our result with the given options:

(A) $x^{-2}$ (Incorrect)

(B) $-x^{-2}$ (Correct)

(C) $1/x^2$ (Incorrect - missing the negative sign)

(D) $-\frac{1}{x^2}$ (Correct)

Both options (B) and (D) represent the correct derivative of $f(x) = \frac{1}{x}$.

The most direct form obtained by applying the power rule is often $-x^{-2}$, which corresponds to option (B).

The equivalent form with a positive exponent, $-\frac{1}{x^2}$, corresponds to option (D).

Therefore, both (B) and (D) are correct answers.

We are asked to evaluate the limit: $\lim\limits_{x \to 1} \frac{\sqrt{x} - 1}{x - 1}$.

First, let's try to substitute $x=1$ directly into the expression:

$\frac{\sqrt{1} - 1}{1 - 1} = \frac{1 - 1}{0} = \frac{0}{0}$.

This is an indeterminate form ($\frac{0}{0}$), which means we need to perform algebraic manipulation on the expression before evaluating the limit.

Consider the denominator $x - 1$. We can rewrite this as a difference of squares by noticing that $x = (\sqrt{x})^2$ and $1 = 1^2$.

So, $x - 1 = (\sqrt{x})^2 - 1^2 = (\sqrt{x} - 1)(\sqrt{x} + 1)$.

Now, substitute this factorization back into the original expression:

$\frac{\sqrt{x} - 1}{x - 1} = \frac{\sqrt{x} - 1}{(\sqrt{x} - 1)(\sqrt{x} + 1)}$.

Since we are evaluating the limit as $x \to 1$, $x$ is approaching 1 but is not equal to 1. Therefore, $\sqrt{x} - 1 \neq 0$, and we can cancel the common factor $(\sqrt{x} - 1)$ from the numerator and the denominator:

$\frac{\cancel{\sqrt{x} - 1}}{\cancel{(\sqrt{x} - 1)}(\sqrt{x} + 1)} = \frac{1}{\sqrt{x} + 1}$ for $x \neq 1$.

Now we can evaluate the limit of the simplified expression by direct substitution:

$\lim\limits_{x \to 1} \frac{1}{\sqrt{x} + 1} = \frac{1}{\sqrt{1} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$.

Alternatively, we could use rationalization. Multiply the numerator and denominator by the conjugate of the numerator, $\sqrt{x} + 1$:

$\lim\limits_{x \to 1} \frac{\sqrt{x} - 1}{x - 1} \times \frac{\sqrt{x} + 1}{\sqrt{x} + 1} = \lim\limits_{x \to 1} \frac{(\sqrt{x})^2 - 1^2}{(x - 1)(\sqrt{x} + 1)}$

$= \lim\limits_{x \to 1} \frac{x - 1}{(x - 1)(\sqrt{x} + 1)}$

For $x \neq 1$, we cancel the $(x-1)$ term:

$= \lim\limits_{x \to 1} \frac{1}{\sqrt{x} + 1}$

Now substitute $x=1$:

$= \frac{1}{\sqrt{1} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$.

Both methods yield the same result.

The value of the limit is $\frac{1}{2}$.

Comparing with the given options, the correct option is (B) 1/2.

We need to find the derivative of the function $f(x) = \cot(4x)$.

This is a composite function, so we will use the Chain Rule.

Let the outer function be $g(u) = \cot(u)$ and the inner function be $u(x) = 4x$.

So, $f(x) = g(u(x)) = \cot(4x)$.

According to the Chain Rule, the derivative of $f(x)$ is given by $f'(x) = g'(u(x)) \cdot u'(x)$.

First, find the derivative of the outer function $g(u) = \cot(u)$ with respect to $u$:

The derivative of $\cot(u)$ is $-\text{cosec}^2(u)$.

So, $g'(u) = -\text{cosec}^2(u)$.

Next, find the derivative of the inner function $u(x) = 4x$ with respect to $x$:

The derivative of $4x$ is 4.

So, $u'(x) = 4$.

Now, apply the Chain Rule formula $f'(x) = g'(u(x)) \cdot u'(x)$.

Substitute $u(x) = 4x$ into $g'(u)$ to get $g'(u(x)) = -\text{cosec}^2(4x)$.

Multiply $g'(u(x))$ by $u'(x) = 4$:

$f'(x) = (-\text{cosec}^2(4x)) \cdot (4)$

$f'(x) = -4 \text{cosec}^2(4x)$.

Thus, the derivative of $f(x) = \cot(4x)$ is $-4 \text{cosec}^2(4x)$.

Comparing our result with the given options:

(A) $-4 \text{cosec}^2(4x)$ (Correct)

(B) $-\text{cosec}^2(4x)$ (Incorrect - missing the factor of 4 from the inner derivative)

(C) $4 \text{cosec}^2(4x)$ (Incorrect - incorrect sign)

(D) $-4 \cot(4x)$ (Incorrect - this is not the derivative)

The correct option is (A) $-4 \text{cosec}^2(4x)$.

We need to evaluate the limit: $\lim\limits_{x \to 0} \frac{e^{3x} - e^{2x}}{x}$.

First, let's try to substitute $x=0$ into the expression:

$\frac{e^{3(0)} - e^{2(0)}}{0} = \frac{e^0 - e^0}{0} = \frac{1 - 1}{0} = \frac{0}{0}$.

This is an indeterminate form ($\frac{0}{0}$), which means we need to use another method to evaluate the limit.

We can use the standard limit formula: $\lim\limits_{u \to 0} \frac{e^u - 1}{u} = 1$.

To apply this, we can rewrite the numerator by subtracting and adding 1:

$\frac{e^{3x} - e^{2x}}{x} = \frac{e^{3x} - 1 - e^{2x} + 1}{x} = \frac{(e^{3x} - 1) - (e^{2x} - 1)}{x}$.

Now, we can split the fraction into two parts:

$\lim\limits_{x \to 0} \frac{(e^{3x} - 1) - (e^{2x} - 1)}{x} = \lim\limits_{x \to 0} \left( \frac{e^{3x} - 1}{x} - \frac{e^{2x} - 1}{x} \right)$.

Consider the first term: $\lim\limits_{x \to 0} \frac{e^{3x} - 1}{x}$. To match the standard form $\frac{e^u - 1}{u}$, let $u = 3x$. As $x \to 0$, $u = 3x \to 0$. We multiply the denominator by 3 and also the expression by 3:

$\lim\limits_{x \to 0} \frac{e^{3x} - 1}{x} = \lim\limits_{x \to 0} \frac{e^{3x} - 1}{3x} \cdot 3$.

Let $u = 3x$. As $x \to 0$, $u \to 0$. So this becomes $3 \lim\limits_{u \to 0} \frac{e^u - 1}{u} = 3 \cdot 1 = 3$.

Consider the second term: $\lim\limits_{x \to 0} \frac{e^{2x} - 1}{x}$. To match the standard form, let $v = 2x$. As $x \to 0$, $v = 2x \to 0$. We multiply the denominator by 2 and also the expression by 2:

$\lim\limits_{x \to 0} \frac{e^{2x} - 1}{x} = \lim\limits_{x \to 0} \frac{e^{2x} - 1}{2x} \cdot 2$.

Let $v = 2x$. As $x \to 0$, $v \to 0$. So this becomes $2 \lim\limits_{v \to 0} \frac{e^v - 1}{v} = 2 \cdot 1 = 2$.

Now, substitute these results back into the limit of the difference:

$\lim\limits_{x \to 0} \left( \frac{e^{3x} - 1}{x} - \frac{e^{2x} - 1}{x} \right) = \lim\limits_{x \to 0} \frac{e^{3x} - 1}{x} - \lim\limits_{x \to 0} \frac{e^{2x} - 1}{x} = 3 - 2 = 1$.

Alternate Solution (Using L'Hopital's Rule):

Since the limit is in the indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule, which states that if $\lim\limits_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim\limits_{x \to c} \frac{f(x)}{g(x)} = \lim\limits_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.

Let $f(x) = e^{3x} - e^{2x}$ and $g(x) = x$.

Find the derivatives of $f(x)$ and $g(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(e^{3x} - e^{2x})$

$f'(x) = \frac{d}{dx}(e^{3x}) - \frac{d}{dx}(e^{2x})$

Using the chain rule, $\frac{d}{dx}(e^{kx}) = k e^{kx}$:

$f'(x) = 3e^{3x} - 2e^{2x}$.

$g'(x) = \frac{d}{dx}(x) = 1$.

Now apply L'Hopital's Rule:

$\lim\limits_{x \to 0} \frac{e^{3x} - e^{2x}}{x} = \lim\limits_{x \to 0} \frac{3e^{3x} - 2e^{2x}}{1}$.

Substitute $x=0$ into the expression:

$\frac{3e^{3(0)} - 2e^{2(0)}}{1} = \frac{3e^0 - 2e^0}{1} = \frac{3(1) - 2(1)}{1} = \frac{3 - 2}{1} = \frac{1}{1} = 1$.

Both methods give the same result, which is 1.

Comparing with the given options:

(A) 1 (Correct)

(B) 2 (Incorrect)

(C) 3 (Incorrect)

(D) $3-2=1$ (Correct, but typically presented as just 1)

The most concise and standard answer is 1, corresponding to option (A).

Let's evaluate the derivative of the function $f(x) = x^2$.

Using the power rule of differentiation, $\frac{d}{dx}(x^n) = nx^{n-1}$.

For $f(x) = x^2$, where $n=2$, the derivative is:

$f'(x) = 2 \cdot x^{2-1} = 2x^1 = 2x$.

Now, let's find the derivative at $x=3$. We substitute $x=3$ into the expression for $f'(x)$:

$f'(3) = 2(3) = 6$.

Let's analyze the given Assertion (A) and Reason (R).

Assertion (A): The derivative of $f(x) = x^2$ at $x=3$ is 6.

From our calculation, the derivative of $f(x)=x^2$ at $x=3$ is indeed 6. So, Assertion (A) is true.

Reason (R): $f'(x) = 2x$, and $f'(3) = 2(3) = 6$.

The statement says that the derivative of $f(x) = x^2$ is $f'(x) = 2x$. This is correct by the power rule.

The statement also says that $f'(3) = 2(3) = 6$. This is the correct evaluation of the derivative at $x=3$.

So, Reason (R) is also true.

Now, let's determine if Reason (R) is the correct explanation for Assertion (A).

Reason (R) provides the general derivative formula for $f(x)=x^2$ and then correctly uses that formula to find the specific derivative value at $x=3$. This is exactly how one would justify or explain why the derivative at $x=3$ is 6.

Therefore, Reason (R) is the correct explanation of Assertion (A).

Based on this analysis, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Comparing with the given options:

(A) Both A and R are true and R is the correct explanation of A. (Correct)

(B) Both A and R are true but R is not the correct explanation of A. (Incorrect)

(C) A is true but R is false. (Incorrect)

(D) A is false but R is true. (Incorrect)

The correct option is (A) Both A and R are true and R is the correct explanation of A.

We are asked to find the derivative of the function $f(x) = x^{-3}$.

To find the derivative of a function of the form $x^n$, we use the power rule of differentiation.

The power rule states that if $f(x) = x^n$, where $n$ is a real number, then its derivative is given by $f'(x) = nx^{n-1}$.

In our function, $f(x) = x^{-3}$, the exponent is $n = -3$.

Applying the power rule, we get:

$f'(x) = n \cdot x^{n-1}$

$f'(x) = (-3) \cdot x^{(-3)-1}$

$f'(x) = -3 \cdot x^{-4}$

$f'(x) = -3x^{-4}$.

Thus, the derivative of $f(x) = x^{-3}$ is $-3x^{-4}$.

Comparing our result with the given options:

(A) $-3x^{-4}$ (Correct)

(B) $-3x^{-2}$ (Incorrect - the exponent should be $n-1 = -3-1 = -4$, not $-2$)

(C) $-3x^4$ (Incorrect - the exponent should be negative, and it should be $-4$)

(D) $x^{-4}$ (Incorrect - missing the coefficient $-3$)

The correct option is (A) $-3x^{-4}$.

We are asked to find the derivative of the function $f(x) = \log_e |x|$ for $x \neq 0$.

We know the definition of the absolute value function:

$|x| = x$ for $x > 0$

$|x| = -x$ for $x < 0$

We need to consider two cases based on the value of $x$ since the domain is $x \neq 0$.

Case 1: $x > 0$

If $x > 0$, then $|x| = x$.

The function becomes $f(x) = \log_e x$.

The derivative of $\log_e x$ with respect to $x$ is $\frac{1}{x}$.

So, for $x > 0$, $f'(x) = \frac{1}{x}$.

Case 2: $x < 0$

If $x < 0$, then $|x| = -x$.

The function becomes $f(x) = \log_e (-x)$.

This is a composite function, so we use the Chain Rule.

Let $u = -x$. Then $f(x) = \log_e u$.

The derivative with respect to $x$ is $f'(x) = \frac{d}{du}(\log_e u) \cdot \frac{du}{dx}$.

The derivative of $\log_e u$ with respect to $u$ is $\frac{1}{u}$.

The derivative of $u = -x$ with respect to $x$ is $\frac{d}{dx}(-x) = -1$.

Substituting these into the Chain Rule formula:

$f'(x) = \frac{1}{u} \cdot (-1)$.

Now, substitute $u = -x$ back:

$f'(x) = \frac{1}{-x} \cdot (-1) = \frac{-1}{-x} = \frac{1}{x}$.

So, for $x < 0$, $f'(x) = \frac{1}{x}$.

Combining both cases, we see that for $x \neq 0$, the derivative of $f(x) = \log_e |x|$ is $\frac{1}{x}$.

$f'(x) = \frac{1}{x}$ for $x \neq 0$.

Comparing our result with the given options:

(A) $1/x$ (Correct)

(B) $1/|x|$ (Incorrect - while this is numerically equal to $1/x$ for $x>0$, it is $\frac{1}{-x}$ for $x<0$, which is not equal to $1/x$ for $x<0$. However, the derivative is $1/x$ for both cases).

(C) $\log_e x$ (Incorrect - this is the function itself, not its derivative)

(D) $1/(x|x|)$ (Incorrect)

The correct option is (A) $1/x$.

We need to evaluate the limit: $\lim\limits_{x \to 0} \frac{\sin x \cos x}{x}$.

First, let's substitute $x=0$ into the expression:

$\frac{\sin(0) \cos(0)}{0} = \frac{0 \cdot 1}{0} = \frac{0}{0}$.

This is an indeterminate form ($\frac{0}{0}$), so we need to use a different method.

We can rewrite the expression using the properties of limits and the standard limit $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$.

The expression can be written as the product of two terms: $\frac{\sin x}{x}$ and $\cos x$.

$\frac{\sin x \cos x}{x} = \left(\frac{\sin x}{x}\right) \cdot \cos x$.

Now, we can apply the limit to each term separately:

$\lim\limits_{x \to 0} \frac{\sin x \cos x}{x} = \lim\limits_{x \to 0} \left(\frac{\sin x}{x}\right) \cdot \left(\lim\limits_{x \to 0} \cos x\right)$.

We know the standard limit:

$\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$.

And the limit of $\cos x$ as $x \to 0$ is obtained by direct substitution:

$\lim\limits_{x \to 0} \cos x = \cos(0) = 1$.

Multiply the results of the two limits:

$\left(\lim\limits_{x \to 0} \frac{\sin x}{x}\right) \cdot \left(\lim\limits_{x \to 0} \cos x\right) = 1 \cdot 1 = 1$.

Alternate Solution (Using L'Hopital's Rule):

Since the limit is of the indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule. We need to find the derivatives of the numerator and the denominator.

Let $f(x) = \sin x \cos x$ and $g(x) = x$.

The derivative of $g(x) = x$ is $g'(x) = 1$.

The derivative of $f(x) = \sin x \cos x$ requires the product rule: $\frac{d}{dx}(uv) = u'v + uv'$.

Let $u = \sin x$ and $v = \cos x$. Then $u' = \cos x$ and $v' = -\sin x$.

$f'(x) = (\cos x)(\cos x) + (\sin x)(-\sin x) = \cos^2 x - \sin^2 x$.

Using the double angle identity, $\cos^2 x - \sin^2 x = \cos(2x)$. So, $f'(x) = \cos(2x)$.

Applying L'Hopital's Rule:

$\lim\limits_{x \to 0} \frac{\sin x \cos x}{x} = \lim\limits_{x \to 0} \frac{f'(x)}{g'(x)} = \lim\limits_{x \to 0} \frac{\cos(2x)}{1}$.

Substitute $x=0$ into the derivative expression:

$\frac{\cos(2 \cdot 0)}{1} = \frac{\cos(0)}{1} = \frac{1}{1} = 1$.

Both methods yield the same result.

The value of the limit is 1.

Comparing with the given options:

(A) 0 (Incorrect)

(B) 1 (Correct)

(C) 2 (Incorrect)

(D) 1/2 (Incorrect)

The correct option is (B) 1.

We need to find the derivative of the function $f(x) = e^{-x}$.

This is a composite function, where an exponential function has another function as its exponent. We will use the Chain Rule.

The Chain Rule states that if $f(x) = g(u(x))$, then $f'(x) = g'(u(x)) \cdot u'(x)$.

In this case, let the outer function be $g(u) = e^u$ and the inner function be $u(x) = -x$.

So, $f(x) = g(u(x)) = e^{-x}$.

First, find the derivative of the outer function $g(u) = e^u$ with respect to $u$:

The derivative of $e^u$ is $e^u$.

So, $g'(u) = e^u$.

Next, find the derivative of the inner function $u(x) = -x$ with respect to $x$:

The derivative of $-x$ is $-1$.

So, $u'(x) = -1$.

Now, apply the Chain Rule formula $f'(x) = g'(u(x)) \cdot u'(x)$.

Substitute $u(x) = -x$ back into $g'(u)$: $g'(u(x)) = e^{-x}$.

Multiply $g'(u(x))$ by $u'(x) = -1$:

$f'(x) = (e^{-x}) \cdot (-1)$

$f'(x) = -e^{-x}$.

Thus, the derivative of $f(x) = e^{-x}$ is $-e^{-x}$.

Comparing our result with the given options:

(A) $e^{-x}$ (Incorrect - missing the negative sign from the inner derivative)

(B) $-e^{-x}$ (Correct)

(C) $e^{x}$ (Incorrect - incorrect exponent and sign)

(D) $-e^{x}$ (Incorrect - incorrect exponent)

The correct option is (B) $-e^{-x}$.

We need to evaluate the limit $\lim\limits_{x \to 0} \frac{1 - \cos^2 x}{x^2}$.

We use the trigonometric identity $\sin^2 x + \cos^2 x = 1$.

From this identity, we can write $1 - \cos^2 x = \sin^2 x$.

Substitute this into the limit expression:

$\lim\limits_{x \to 0} \frac{1 - \cos^2 x}{x^2} = \lim\limits_{x \to 0} \frac{\sin^2 x}{x^2}$

We can rewrite $\frac{\sin^2 x}{x^2}$ as $\left(\frac{\sin x}{x}\right)^2$.

So, the limit becomes:

$\lim\limits_{x \to 0} \left(\frac{\sin x}{x}\right)^2$

We know the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

Using the property that $\lim\limits_{x \to a} [f(x)]^n = [\lim\limits_{x \to a} f(x)]^n$, we can apply the limit inside the square:

$\lim\limits_{x \to 0} \left(\frac{\sin x}{x}\right)^2 = \left(\lim\limits_{x \to 0} \frac{\sin x}{x}\right)^2$

Substitute the value of the standard limit:

$\left(\lim\limits_{x \to 0} \frac{\sin x}{x}\right)^2 = (1)^2 = 1$

The value of the limit is 1.

Comparing this result with the given options, we see that option (B) matches the calculated value.

The final answer is $\textbf{1}$.

The correct option is (B).

We are asked to evaluate the limit of the polynomial function $(3x^2 - x + 1)$ as $x$ approaches $2$.

The limit of a polynomial function can be found by direct substitution. If $P(x)$ is a polynomial and $a$ is any real number, then $\lim\limits_{x \to a} P(x) = P(a)$.

In this problem, the polynomial function is $P(x) = 3x^2 - x + 1$ and the value $x$ approaches is $a = 2$.

So, we substitute $x=2$ into the expression:

$\lim\limits_{x \to 2} (3x^2 - x + 1) = 3(2)^2 - (2) + 1$

Now, we perform the arithmetic operations:

$3(2)^2 - 2 + 1 = 3(4) - 2 + 1$

$= 12 - 2 + 1$

$= 10 + 1$

$= 11$

Thus, the value of the limit is $\mathbf{11}$.

The final evaluated limit is:

$\lim\limits_{x \to 2} (3x^2 - x + 1) = \textbf{11}$

We are asked to evaluate the limit $\lim\limits_{x \to 0} \frac{(x+1)^5 - 1}{x}$.

If we substitute $x=0$ directly into the expression, we get $\frac{(0+1)^5 - 1}{0} = \frac{1^5 - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0}$, which is an indeterminate form.

We can evaluate this limit using a standard limit formula, which is $\lim\limits_{u \to a} \frac{u^n - a^n}{u - a} = na^{n-1}$.

Let $u = x+1$. As $x \to 0$, $u = 0+1 = 1$. So, $u \to 1$.

Also, $x = u-1$.

Substituting these into the given limit, we get:

$\lim\limits_{x \to 0} \frac{(x+1)^5 - 1}{x} = \lim\limits_{u \to 1} \frac{u^5 - 1}{u - 1}$

This new limit is in the form $\lim\limits_{u \to a} \frac{u^n - a^n}{u - a}$ with $a=1$ and $n=5$.

Applying the formula, the limit is $na^{n-1} = 5(1)^{5-1} = 5(1)^4$.

$5(1)^4 = 5 \times 1 = 5$.

Therefore, the value of the limit is $\mathbf{5}$.

$\lim\limits_{x \to 0} \frac{(x+1)^5 - 1}{x} = \textbf{5}$

We are asked to evaluate the limit $\lim\limits_{x \to 3} \frac{x^2 - 9}{x - 3}$.

If we attempt direct substitution of $x=3$ into the expression, we get:

$\frac{(3)^2 - 9}{3 - 3} = \frac{9 - 9}{0} = \frac{0}{0}$

This is an indeterminate form, which means we need to simplify the expression before evaluating the limit.

The numerator, $x^2 - 9$, is a difference of squares. It can be factored as $(x-3)(x+3)$.

So, the expression becomes:

$\frac{x^2 - 9}{x - 3} = \frac{(x-3)(x+3)}{x - 3}$

For $x \neq 3$, we can cancel the common factor $(x-3)$ from the numerator and the denominator.

$\frac{\cancel{(x-3)}(x+3)}{\cancel{(x-3)}} = x+3 \quad (\text{for } x \neq 3)$

Since the limit is concerned with values of $x$ approaching $3$ but not equal to $3$, the expression $\frac{x^2 - 9}{x - 3}$ is equivalent to $x+3$ in the neighborhood of $x=3$.

Therefore, the limit can be rewritten as:

$\lim\limits_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim\limits_{x \to 3} (x+3)$

Now, we can evaluate this limit by direct substitution, as $x+3$ is a polynomial function.

Substitute $x=3$ into the simplified expression:

$\lim\limits_{x \to 3} (x+3) = 3 + 3 = 6$

Thus, the value of the limit is $\mathbf{6}$.

$\lim\limits_{x \to 3} \frac{x^2 - 9}{x - 3} = \textbf{6}$

We need to evaluate the limit $\lim\limits_{x \to 0} \frac{\sin 4x}{\sin 2x}$.

Upon substituting $x=0$ into the expression, we get $\frac{\sin(4 \times 0)}{\sin(2 \times 0)} = \frac{\sin 0}{\sin 0} = \frac{0}{0}$, which is an indeterminate form.

We can use the standard trigonometric limit formula: $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

We can rewrite the given expression by multiplying and dividing the numerator by $4x$ and the denominator by $2x$:

$\frac{\sin 4x}{\sin 2x} = \frac{\frac{\sin 4x}{4x} \times 4x}{\frac{\sin 2x}{2x} \times 2x}$

This can be rearranged as:

$\frac{\sin 4x}{\sin 2x} = \frac{\frac{\sin 4x}{4x}}{\frac{\sin 2x}{2x}} \times \frac{4x}{2x}$

For $x \neq 0$, the term $\frac{4x}{2x}$ simplifies to $2$.

So, $\frac{\sin 4x}{\sin 2x} = \frac{\frac{\sin 4x}{4x}}{\frac{\sin 2x}{2x}} \times 2$

Now, we can take the limit as $x \to 0$. As $x \to 0$, $4x \to 0$ and $2x \to 0$.

Using the limit properties $\lim\limits_{x \to a} (c \cdot f(x)) = c \cdot \lim\limits_{x \to a} f(x)$ and $\lim\limits_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim\limits_{x \to a} f(x)}{\lim\limits_{x \to a} g(x)}$ (provided the limits exist and the denominator limit is non-zero), we have:

$\lim\limits_{x \to 0} \frac{\sin 4x}{\sin 2x} = \lim\limits_{x \to 0} \left( 2 \times \frac{\frac{\sin 4x}{4x}}{\frac{\sin 2x}{2x}} \right)$

$= 2 \times \frac{\lim\limits_{x \to 0} \frac{\sin 4x}{4x}}{\lim\limits_{x \to 0} \frac{\sin 2x}{2x}}$

Applying the standard limit formula $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ for $\theta=4x$ and $\theta=2x$:

$\lim\limits_{x \to 0} \frac{\sin 4x}{4x} = 1$

$\lim\limits_{x \to 0} \frac{\sin 2x}{2x} = 1$

Substituting these values into the limit expression:

$= 2 \times \frac{1}{1}$

$= 2 \times 1$

$= 2$

Therefore, the value of the limit is $\mathbf{2}$.

$\lim\limits_{x \to 0} \frac{\sin 4x}{\sin 2x} = \textbf{2}$

We want to evaluate the limit $\lim\limits_{x \to 0} \frac{\tan ax}{bx}$.

As $x \to 0$, the numerator $\tan(ax) \to \tan(a \cdot 0) = \tan(0) = 0$, and the denominator $bx \to b \cdot 0 = 0$. This limit is in the indeterminate form $\frac{0}{0}$.

We can use the standard limit formula $\lim\limits_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$.

To apply this formula, we need the argument of the tangent ($ax$) in the denominator. We can achieve this by multiplying and dividing the expression by $ax$, assuming $a \neq 0$ and $b \neq 0$ and $x \neq 0$ near the limit.

Consider the expression $\frac{\tan ax}{bx}$ for $x \neq 0$:

$\frac{\tan ax}{bx} = \frac{\tan ax}{bx} \times \frac{ax}{ax}$

Rearranging the terms, we get:

$\frac{\tan ax}{bx} = \frac{\tan ax}{ax} \times \frac{ax}{bx}$

Now, we can simplify the second fraction:

$\frac{ax}{bx} = \frac{\cancel{x} \cdot a}{\cancel{x} \cdot b} = \frac{a}{b}$ (assuming $x \neq 0$ and $b \neq 0$)

So, the expression becomes:

$\frac{\tan ax}{bx} = \frac{\tan ax}{ax} \times \frac{a}{b}$

Now, we can apply the limit as $x \to 0$:

$\lim\limits_{x \to 0} \frac{\tan ax}{bx} = \lim\limits_{x \to 0} \left( \frac{\tan ax}{ax} \times \frac{a}{b} \right)$

Using the property that the limit of a product is the product of the limits:

$\lim\limits_{x \to 0} \frac{\tan ax}{bx} = \left( \lim\limits_{x \to 0} \frac{\tan ax}{ax} \right) \times \left( \lim\limits_{x \to 0} \frac{a}{b} \right)$

For the first limit, let $\theta = ax$. As $x \to 0$, $ax \to a \cdot 0 = 0$ (assuming $a$ is finite). So, $\theta \to 0$. Using the standard limit:

$\lim\limits_{x \to 0} \frac{\tan ax}{ax} = \lim\limits_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$

For the second limit, $\frac{a}{b}$ is a constant with respect to $x$. The limit of a constant is the constant itself:

$\lim\limits_{x \to 0} \frac{a}{b} = \frac{a}{b}$

Combining the results:

$\lim\limits_{x \to 0} \frac{\tan ax}{bx} = (1) \times \left( \frac{a}{b} \right)$

$\lim\limits_{x \to 0} \frac{\tan ax}{bx} = \frac{a}{b}$

This result holds provided $b \neq 0$. If $b=0$, the original expression is undefined unless $a=0$. If $a \neq 0, b=0$, the limit does not exist. If $a=0, b=0$, the expression is $0/0$ but the original function is $\frac{\tan 0}{0x}$ which is not defined in an interval around 0.

Assuming $b \neq 0$, the value of the limit is $\frac{a}{b}$.

The given function is $f(x) = 3x^2 - 5x + 4$.

The derivative of a function $f(x)$ using the first principle (definition of derivative) is given by:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

First, we find $f(x+h)$ by substituting $(x+h)$ for $x$ in the expression for $f(x)$:

$f(x+h) = 3(x+h)^2 - 5(x+h) + 4$

$f(x+h) = 3(x^2 + 2xh + h^2) - 5x - 5h + 4$

$f(x+h) = 3x^2 + 6xh + 3h^2 - 5x - 5h + 4$

Next, we find the difference $f(x+h) - f(x)$:

$f(x+h) - f(x) = (3x^2 + 6xh + 3h^2 - 5x - 5h + 4) - (3x^2 - 5x + 4)$

$f(x+h) - f(x) = 3x^2 + 6xh + 3h^2 - 5x - 5h + 4 - 3x^2 + 5x - 4$

Combining like terms:

$f(x+h) - f(x) = (3x^2 - 3x^2) + (-5x + 5x) + (4 - 4) + 6xh + 3h^2 - 5h$

$f(x+h) - f(x) = 6xh + 3h^2 - 5h$

Now, we form the difference quotient $\frac{f(x+h) - f(x)}{h}$:

$\frac{f(x+h) - f(x)}{h} = \frac{6xh + 3h^2 - 5h}{h}$

Factor out $h$ from the numerator:

$\frac{h(6x + 3h - 5)}{h}$

For $h \neq 0$, we can cancel the $h$ terms:

$\frac{f(x+h) - f(x)}{h} = 6x + 3h - 5$

Finally, we evaluate the limit of this expression as $h \to 0$:

$f'(x) = \lim\limits_{h \to 0} (6x + 3h - 5)$

As $h$ approaches $0$, the term $3h$ approaches $3 \times 0 = 0$. The terms $6x$ and $-5$ do not depend on $h$.

$f'(x) = 6x + 0 - 5$

$f'(x) = 6x - 5$

Thus, the derivative of $f(x) = 3x^2 - 5x + 4$ using the first principle is $\mathbf{6x - 5}$.

The given function is $f(x) = \sin x$.

The derivative of a function $f(x)$ using the first principle (definition of derivative) is given by:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

We need to find $f(x+h)$ and $f(x+h) - f(x)$.

$f(x+h) = \sin(x+h)$

$f(x+h) - f(x) = \sin(x+h) - \sin x$

Now substitute this into the limit formula:

$f'(x) = \lim\limits_{h \to 0} \frac{\sin(x+h) - \sin x}{h}$

We use the trigonometric identity for the difference of sines: $\sin A - \sin B = 2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$.

Let $A = x+h$ and $B = x$.

Then $\frac{A+B}{2} = \frac{(x+h) + x}{2} = \frac{2x+h}{2} = x + \frac{h}{2}$.

And $\frac{A-B}{2} = \frac{(x+h) - x}{2} = \frac{h}{2}$.

Substituting these into the numerator:

$\sin(x+h) - \sin x = 2 \cos \left(x + \frac{h}{2}\right) \sin \left(\frac{h}{2}\right)$

So the limit becomes:

$f'(x) = \lim\limits_{h \to 0} \frac{2 \cos \left(x + \frac{h}{2}\right) \sin \left(\frac{h}{2}\right)}{h}$

We can rearrange the terms to make use of the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

$f'(x) = \lim\limits_{h \to 0} \left[ \cos \left(x + \frac{h}{2}\right) \cdot \frac{2 \sin \left(\frac{h}{2}\right)}{h} \right]$

$f'(x) = \lim\limits_{h \to 0} \left[ \cos \left(x + \frac{h}{2}\right) \cdot \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}} \right]$

Using the property that the limit of a product is the product of the limits:

$f'(x) = \left( \lim\limits_{h \to 0} \cos \left(x + \frac{h}{2}\right) \right) \cdot \left( \lim\limits_{h \to 0} \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}} \right)$

Evaluate the first limit:

As $h \to 0$, $\frac{h}{2} \to 0$. By the continuity of the cosine function, $\lim\limits_{h \to 0} \cos \left(x + \frac{h}{2}\right) = \cos(x+0) = \cos x$.

Evaluate the second limit:

Let $\theta = \frac{h}{2}$. As $h \to 0$, $\theta \to 0$. Using the standard limit, $\lim\limits_{h \to 0} \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}} = \lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

Substitute the values of the limits back:

$f'(x) = (\cos x) \cdot (1)$

$f'(x) = \cos x$

Thus, the derivative of $f(x) = \sin x$ using the first principle is $\mathbf{\cos x}$.

The given function is $f(x) = x^3 - 2x$. We need to find its derivative using the power rule.

The power rule for differentiation states that for a function $f(x) = x^n$, its derivative is $f'(x) = nx^{n-1}$, where $n$ is any real number.

The derivative of a constant multiple of a function is given by $\frac{d}{dx}[cf(x)] = c \frac{d}{dx}[f(x)]$, where $c$ is a constant.

The derivative of a sum or difference of functions is the sum or difference of their derivatives: $\frac{d}{dx}[f(x) \pm g(x)] = \frac{d}{dx}[f(x)] \pm \frac{d}{dx}[g(x)]$.

We can find the derivative of each term in the function $f(x) = x^3 - 2x$ separately.

For the first term, $x^3$, we apply the power rule with $n=3$:

$\frac{d}{dx}(x^3) = 3x^{3-1} = 3x^2$

For the second term, $-2x$, which can be written as $-2 \times x^1$, we apply the constant multiple rule and the power rule with $c=-2$ and $n=1$:

$\frac{d}{dx}(-2x) = -2 \frac{d}{dx}(x^1) = -2 \times 1 \times x^{1-1} = -2x^0$

Since $x^0 = 1$ for $x \neq 0$, this simplifies to $-2 \times 1 = -2$.

$\frac{d}{dx}(-2x) = -2$

Now, we combine the derivatives of the terms using the difference rule:

$f'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(2x)$

$f'(x) = 3x^2 - 2$

Thus, the derivative of $f(x) = x^3 - 2x$ using the power rule is $\mathbf{3x^2 - 2}$.

The given function is $f(x) = \frac{x+1}{x-1}$. We will use the quotient rule to find its derivative.

Let the numerator be $u(x) = x+1$ and the denominator be $v(x) = x-1$.

The quotient rule for differentiation states that if $f(x) = \frac{u(x)}{v(x)}$, then its derivative $f'(x)$ is given by:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

First, we find the derivatives of $u(x)$ and $v(x)$.

Using the power rule and the rule for the derivative of a constant:

$u(x) = x+1$

$u'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(1) = 1x^{1-1} + 0 = 1 \times 1 + 0 = 1$

So, $u'(x) = 1$.

$v(x) = x-1$

$v'(x) = \frac{d}{dx}(x) - \frac{d}{dx}(1) = 1x^{1-1} - 0 = 1 \times 1 - 0 = 1$

So, $v'(x) = 1$.

Now, substitute $u(x)$, $v(x)$, $u'(x)$, and $v'(x)$ into the quotient rule formula:

$f'(x) = \frac{(1)(x-1) - (x+1)(1)}{(x-1)^2}$

Simplify the numerator:

Numerator $= (x-1) - (x+1)$

$= x - 1 - x - 1$

$= (x-x) + (-1-1)$

$= 0 - 2$

$= -2$

So, the derivative is:

$f'(x) = \frac{-2}{(x-1)^2}$

Thus, the derivative of $f(x) = \frac{x+1}{x-1}$ using the quotient rule is $\mathbf{f'(x) = \frac{-2}{(x-1)^2}}$.

We are asked to evaluate the limit $\lim\limits_{x \to 1} \frac{x^{10} - 1}{x - 1}$.

If we attempt direct substitution of $x=1$, we get:

$\frac{(1)^{10} - 1}{1 - 1} = \frac{1 - 1}{0} = \frac{0}{0}$

This is an indeterminate form.

This limit is in the standard form $\lim\limits_{x \to a} \frac{x^n - a^n}{x - a}$.

The value of this standard limit is given by the formula $na^{n-1}$.

In our given limit, $\lim\limits_{x \to 1} \frac{x^{10} - 1^ {10}}{x - 1}$, we can identify the parameters by comparing it with the standard form:

The variable is $x$.

The value $x$ approaches is $a = 1$.

The power is $n = 10$.

Applying the formula $na^{n-1}$, we substitute the values $n=10$ and $a=1$:

Limit $= 10 \times (1)^{10-1}$

Limit $= 10 \times (1)^9$

Limit $= 10 \times 1$

Limit $= 10$

Thus, the value of the limit is $\mathbf{10}$.

$\lim\limits_{x \to 1} \frac{x^{10} - 1}{x - 1} = \textbf{10}$

Given limit:

$\lim\limits_{x \to 0} \frac{1 - \cos x}{x^2}$

When $x \to 0$, the numerator $1 - \cos x \to 1 - \cos 0 = 1 - 1 = 0$ and the denominator $x^2 \to 0^2 = 0$. This is an indeterminate form of type $\frac{0}{0}$.

We can evaluate this limit using the standard limit $\lim\limits_{y \to 0} \frac{\sin y}{y} = 1$.

Recall the trigonometric identity: $\cos x = 1 - 2 \sin^2 \left(\frac{x}{2}\right)$.

Substitute this into the expression:

$\frac{1 - \cos x}{x^2} = \frac{1 - \left(1 - 2 \sin^2 \left(\frac{x}{2}\right)\right)}{x^2}$

$= \frac{1 - 1 + 2 \sin^2 \left(\frac{x}{2}\right)}{x^2}$

$= \frac{2 \sin^2 \left(\frac{x}{2}\right)}{x^2}$

Rewrite the expression to use the standard limit form:

$\frac{2 \sin^2 \left(\frac{x}{2}\right)}{x^2} = 2 \cdot \frac{\sin^2 \left(\frac{x}{2}\right)}{x^2}$

$= 2 \left( \frac{\sin \left(\frac{x}{2}\right)}{x} \right)^2$

We want $\frac{\sin y}{y}$ where $y = \frac{x}{2}$. So we need $\frac{x}{2}$ in the denominator.

$= 2 \left( \frac{\sin \left(\frac{x}{2}\right)}{2 \cdot \left(\frac{x}{2}\right)} \right)^2$

$= 2 \left( \frac{1}{2} \cdot \frac{\sin \left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)} \right)^2$

$= 2 \cdot \left(\frac{1}{2}\right)^2 \cdot \left( \frac{\sin \left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)} \right)^2$

$= 2 \cdot \frac{1}{4} \cdot \left( \frac{\sin \left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)} \right)^2$

$= \frac{1}{2} \left( \frac{\sin \left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)} \right)^2$

Now, evaluate the limit:

$\lim\limits_{x \to 0} \frac{1 - \cos x}{x^2} = \lim\limits_{x \to 0} \frac{1}{2} \left( \frac{\sin \left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)} \right)^2$

As $x \to 0$, $\frac{x}{2} \to 0$. Let $y = \frac{x}{2}$. Then as $x \to 0$, $y \to 0$.

So, $\lim\limits_{x \to 0} \frac{\sin \left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)} = \lim\limits_{y \to 0} \frac{\sin y}{y} = 1$.

Therefore,

$\lim\limits_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \cdot (1)^2$

$= \frac{1}{2}$

The final answer is $\frac{1}{2}$.

Alternate Solution using L'Hopital's Rule:

The limit $\lim\limits_{x \to 0} \frac{1 - \cos x}{x^2}$ is of the indeterminate form $\frac{0}{0}$. We can apply L'Hopital's Rule.

Applying L'Hopital's Rule for the first time:

$\lim\limits_{x \to 0} \frac{\frac{d}{dx}(1 - \cos x)}{\frac{d}{dx}(x^2)} = \lim\limits_{x \to 0} \frac{\sin x}{2x}$

This is again of the indeterminate form $\frac{0}{0}$. Apply L'Hopital's Rule again.

Applying L'Hopital's Rule for the second time:

$\lim\limits_{x \to 0} \frac{\frac{d}{dx}(\sin x)}{\frac{d}{dx}(2x)} = \lim\limits_{x \to 0} \frac{\cos x}{2}$

Now, substitute $x = 0$:

$\lim\limits_{x \to 0} \frac{\cos x}{2} = \frac{\cos 0}{2} = \frac{1}{2}$

Both methods yield the same result.

Given limit:

$\lim\limits_{x \to \infty} \frac{3x^2 + 5x + 2}{2x^2 - x + 1}$

To evaluate the limit of a rational function as $x \to \infty$, we can divide both the numerator and the denominator by the highest power of $x$ present in the denominator. In this case, the highest power is $x^2$.

Divide the numerator and the denominator by $x^2$:

$\frac{3x^2 + 5x + 2}{2x^2 - x + 1} = \frac{\frac{3x^2 + 5x + 2}{x^2}}{\frac{2x^2 - x + 1}{x^2}}$

$= \frac{\frac{3x^2}{x^2} + \frac{5x}{x^2} + \frac{2}{x^2}}{\frac{2x^2}{x^2} - \frac{x}{x^2} + \frac{1}{x^2}}$

$= \frac{3 + \frac{5}{x} + \frac{2}{x^2}}{2 - \frac{1}{x} + \frac{1}{x^2}}$

Now, evaluate the limit as $x \to \infty$:

$\lim\limits_{x \to \infty} \frac{3 + \frac{5}{x} + \frac{2}{x^2}}{2 - \frac{1}{x} + \frac{1}{x^2}}$

Using the property $\lim\limits_{x \to \infty} \frac{c}{x^n} = 0$ for any constant $c$ and $n > 0$, we have:

$\lim\limits_{x \to \infty} \frac{5}{x} = 0$

$\lim\limits_{x \to \infty} \frac{2}{x^2} = 0$

$\lim\limits_{x \to \infty} \frac{1}{x} = 0$

$\lim\limits_{x \to \infty} \frac{1}{x^2} = 0$

Substitute these values into the limit expression:

$\lim\limits_{x \to \infty} \frac{3 + \frac{5}{x} + \frac{2}{x^2}}{2 - \frac{1}{x} + \frac{1}{x^2}} = \frac{3 + 0 + 0}{2 - 0 + 0}$

$= \frac{3}{2}$

The final answer is $\frac{3}{2}$.

Given function:

$f(x) = x \sin x$

We need to find the derivative of $f(x)$ using the product rule.

The product rule for differentiation states that if $h(x) = u(x)v(x)$, then the derivative of $h(x)$ is given by $h'(x) = u'(x)v(x) + u(x)v'(x)$.

In our function $f(x) = x \sin x$, let's identify $u(x)$ and $v(x)$.

Let $u(x) = x$ and $v(x) = \sin x$.

Now, find the derivatives of $u(x)$ and $v(x)$ with respect to $x$:

Derivative of $u(x)$:

$u'(x) = \frac{d}{dx}(x)$

$u'(x) = 1$

Derivative of $v(x)$:

$v'(x) = \frac{d}{dx}(\sin x)$

$v'(x) = \cos x$

Now, apply the product rule $f'(x) = u'(x)v(x) + u(x)v'(x)$:

$f'(x) = (1)(\sin x) + (x)(\cos x)$

Simplify the expression:

$f'(x) = \sin x + x \cos x$

The derivative of $f(x) = x \sin x$ is $f'(x) = \sin x + x \cos x$.

Given function:

$f(x) = \tan x$

We need to find the derivative of $f(x) = \tan x$ using the first principle, which is given by the formula:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

First, find $f(x+h)$:

$f(x+h) = \tan(x+h)$

Now, find the difference $f(x+h) - f(x)$:

$f(x+h) - f(x) = \tan(x+h) - \tan x$

Rewrite $\tan$ in terms of $\sin$ and $\cos$:

$= \frac{\sin(x+h)}{\cos(x+h)} - \frac{\sin x}{\cos x}$

Combine the fractions with a common denominator:

$= \frac{\sin(x+h)\cos x - \cos(x+h)\sin x}{\cos(x+h)\cos x}$

Use the trigonometric identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$. Let $A = x+h$ and $B = x$.

So, $\sin((x+h) - x) = \sin(x+h)\cos x - \cos(x+h)\sin x$

$\sin h = \sin(x+h)\cos x - \cos(x+h)\sin x$

Substitute this back into the expression for $f(x+h) - f(x)$:

$f(x+h) - f(x) = \frac{\sin h}{\cos(x+h)\cos x}$

Now, form the difference quotient $\frac{f(x+h) - f(x)}{h}$:

$\frac{f(x+h) - f(x)}{h} = \frac{\frac{\sin h}{\cos(x+h)\cos x}}{h}$

$= \frac{\sin h}{h \cdot \cos(x+h)\cos x}$

$= \frac{\sin h}{h} \cdot \frac{1}{\cos(x+h)\cos x}$

Now, evaluate the limit as $h \to 0$:

$f'(x) = \lim\limits_{h \to 0} \left( \frac{\sin h}{h} \cdot \frac{1}{\cos(x+h)\cos x} \right)$

Using the property of limits of a product, we can write:

$f'(x) = \left(\lim\limits_{h \to 0} \frac{\sin h}{h}\right) \cdot \left(\lim\limits_{h \to 0} \frac{1}{\cos(x+h)\cos x}\right)$

We know the standard limit $\lim\limits_{h \to 0} \frac{\sin h}{h} = 1$.

For the second limit, substitute $h = 0$ (since $\cos x$ is continuous):

$\lim\limits_{h \to 0} \frac{1}{\cos(x+h)\cos x} = \frac{1}{\cos(x+0)\cos x} = \frac{1}{\cos x \cdot \cos x} = \frac{1}{\cos^2 x}$

Combine the results:

$f'(x) = 1 \cdot \frac{1}{\cos^2 x}$

$f'(x) = \frac{1}{\cos^2 x}$

Recall that $\frac{1}{\cos x} = \sec x$. Therefore, $\frac{1}{\cos^2 x} = (\sec x)^2 = \sec^2 x$.

Thus, the derivative of $f(x) = \tan x$ is $f'(x) = \sec^2 x$.

Given limit:

$\lim\limits_{x \to 2} \frac{x^3 - 8}{x^2 - 4}$

As $x \to 2$, the numerator $x^3 - 8 \to 2^3 - 8 = 8 - 8 = 0$.

As $x \to 2$, the denominator $x^2 - 4 \to 2^2 - 4 = 4 - 4 = 0$.

The limit is of the indeterminate form $\frac{0}{0}$.

Method 1: Factoring

Factor the numerator and the denominator.

Numerator: $x^3 - 8 = x^3 - 2^3$. This is a difference of cubes, $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$.

$x^3 - 8 = (x-2)(x^2 + 2x + 2^2) = (x-2)(x^2 + 2x + 4)$.

Denominator: $x^2 - 4 = x^2 - 2^2$. This is a difference of squares, $a^2 - b^2 = (a-b)(a+b)$.

$x^2 - 4 = (x-2)(x+2)$.

Substitute the factored forms into the limit expression:

$\lim\limits_{x \to 2} \frac{(x-2)(x^2 + 2x + 4)}{(x-2)(x+2)}$

Since $x \to 2$, $x \neq 2$, which means $x-2 \neq 0$. We can cancel the common factor $(x-2)$ from the numerator and the denominator.

$\lim\limits_{x \to 2} \frac{\cancel{(x-2)}(x^2 + 2x + 4)}{\cancel{(x-2)}(x+2)} = \lim\limits_{x \to 2} \frac{x^2 + 2x + 4}{x+2}$

Now, substitute $x = 2$ into the simplified expression, as the denominator is not zero at $x=2$:

$\frac{2^2 + 2(2) + 4}{2+2} = \frac{4 + 4 + 4}{4} = \frac{12}{4} = 3$

Thus, $\lim\limits_{x \to 2} \frac{x^3 - 8}{x^2 - 4} = 3$.

Alternate Solution: Using L'Hopital's Rule

Since the limit is of the indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule.

L'Hopital's Rule states that if $\lim\limits_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim\limits_{x \to c} \frac{f(x)}{g(x)} = \lim\limits_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.

Let $f(x) = x^3 - 8$ and $g(x) = x^2 - 4$.

Find the derivative of $f(x)$:

$f'(x) = \frac{d}{dx}(x^3 - 8) = 3x^2$

Find the derivative of $g(x)$:

$g'(x) = \frac{d}{dx}(x^2 - 4) = 2x$

Apply L'Hopital's Rule:

$\lim\limits_{x \to 2} \frac{x^3 - 8}{x^2 - 4} = \lim\limits_{x \to 2} \frac{3x^2}{2x}$

Now, substitute $x=2$ into the new expression:

$\frac{3(2)^2}{2(2)} = \frac{3(4)}{4} = \frac{12}{4} = 3$

Both methods confirm that the limit is 3.

Given limit:

$\lim\limits_{x \to 0} \frac{e^{ax} - 1}{x}$

As $x \to 0$, the numerator $e^{ax} - 1 \to e^{a \cdot 0} - 1 = e^0 - 1 = 1 - 1 = 0$.

As $x \to 0$, the denominator $x \to 0$.

The limit is of the indeterminate form $\frac{0}{0}$.

Method 1: Using Standard Limit

We know the standard limit $\lim\limits_{y \to 0} \frac{e^y - 1}{y} = 1$.

Let $y = ax$. As $x \to 0$, $y = ax \to a \cdot 0 = 0$.

Rewrite the given limit expression to match the standard form:

$\lim\limits_{x \to 0} \frac{e^{ax} - 1}{x}$

Multiply and divide the denominator by $a$:

$= \lim\limits_{x \to 0} \frac{e^{ax} - 1}{\frac{1}{a} (ax)}$

$= \lim\limits_{x \to 0} a \cdot \frac{e^{ax} - 1}{ax}$

Now, let $y = ax$. As $x \to 0$, $y \to 0$.

$= a \cdot \lim\limits_{y \to 0} \frac{e^y - 1}{y}$

Using the standard limit $\lim\limits_{y \to 0} \frac{e^y - 1}{y} = 1$:

$= a \cdot 1$

$= a$

Thus, $\lim\limits_{x \to 0} \frac{e^{ax} - 1}{x} = a$.

Alternate Solution: Using L'Hopital's Rule

Since the limit is of the indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule.

Let $f(x) = e^{ax} - 1$ and $g(x) = x$.

Find the derivative of $f(x)$:

$f'(x) = \frac{d}{dx}(e^{ax} - 1) = a e^{ax}$

Find the derivative of $g(x)$:

$g'(x) = \frac{d}{dx}(x) = 1$

Apply L'Hopital's Rule:

$\lim\limits_{x \to 0} \frac{e^{ax} - 1}{x} = \lim\limits_{x \to 0} \frac{f'(x)}{g'(x)} = \lim\limits_{x \to 0} \frac{a e^{ax}}{1}$

$= \lim\limits_{x \to 0} a e^{ax}$

Substitute $x=0$ into the expression:

$= a e^{a \cdot 0}$

$= a e^0$

$= a \cdot 1$

$= a$

Both methods give the same result.

The final answer is $a$.

Given function:

$f(x) = \cos x - 2 \sin x$

We need to find the derivative of $f(x)$ with respect to $x$. We can use the rules of differentiation for trigonometric functions and the properties of linearity of differentiation.

The derivative of a sum or difference of functions is the sum or difference of their derivatives:

$f'(x) = \frac{d}{dx}(\cos x - 2 \sin x) = \frac{d}{dx}(\cos x) - \frac{d}{dx}(2 \sin x)$

The derivative of a constant times a function is the constant times the derivative of the function:

$\frac{d}{dx}(2 \sin x) = 2 \frac{d}{dx}(\sin x)$

Recall the standard derivatives of $\cos x$ and $\sin x$:

$\frac{d}{dx}(\cos x) = -\sin x$

$\frac{d}{dx}(\sin x) = \cos x$

Substitute these derivatives back into the expression for $f'(x)$:

$f'(x) = (-\sin x) - 2 (\cos x)$

Simplify the expression:

$f'(x) = -\sin x - 2 \cos x$

The derivative of $f(x) = \cos x - 2 \sin x$ is $f'(x) = -\sin x - 2 \cos x$.

Given limit:

$\lim\limits_{h \to 0} \frac{\tan(x+h) - \tan x}{h}$

This limit expression is the definition of the derivative of the function $f(x) = \tan x$ from the first principle.

The derivative of a function $f(x)$ is defined as $f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$.

In this case, $f(x) = \tan x$.

We need to evaluate $\frac{\tan(x+h) - \tan x}{h}$.

First, consider the numerator $\tan(x+h) - \tan x$.

Rewrite $\tan$ in terms of $\sin$ and $\cos$:

$\tan(x+h) - \tan x = \frac{\sin(x+h)}{\cos(x+h)} - \frac{\sin x}{\cos x}$

Find a common denominator and combine the terms:

$= \frac{\sin(x+h)\cos x - \cos(x+h)\sin x}{\cos(x+h)\cos x}$

Use the sine subtraction formula: $\sin(A-B) = \sin A \cos B - \cos A \sin B$.

Let $A = x+h$ and $B = x$. Then $A-B = (x+h) - x = h$.

So, $\sin(x+h)\cos x - \cos(x+h)\sin x = \sin((x+h) - x) = \sin h$.

Substitute this back into the numerator:

$\tan(x+h) - \tan x = \frac{\sin h}{\cos(x+h)\cos x}$

Now, substitute this back into the limit expression:

$\lim\limits_{h \to 0} \frac{\tan(x+h) - \tan x}{h} = \lim\limits_{h \to 0} \frac{\frac{\sin h}{\cos(x+h)\cos x}}{h}$

$= \lim\limits_{h \to 0} \frac{\sin h}{h \cdot \cos(x+h)\cos x}$

We can rewrite this as a product of limits:

$= \lim\limits_{h \to 0} \left( \frac{\sin h}{h} \cdot \frac{1}{\cos(x+h)\cos x} \right)$

$= \left(\lim\limits_{h \to 0} \frac{\sin h}{h}\right) \cdot \left(\lim\limits_{h \to 0} \frac{1}{\cos(x+h)\cos x}\right)$

We use the standard limit $\lim\limits_{h \to 0} \frac{\sin h}{h} = 1$.

For the second limit, as $h \to 0$, $\cos(x+h) \to \cos(x+0) = \cos x$ (since $\cos$ is a continuous function).

So, $\lim\limits_{h \to 0} \frac{1}{\cos(x+h)\cos x} = \frac{1}{\cos x \cdot \cos x} = \frac{1}{\cos^2 x}$.

Combining the results:

$\lim\limits_{h \to 0} \frac{\tan(x+h) - \tan x}{h} = 1 \cdot \frac{1}{\cos^2 x}$

$= \frac{1}{\cos^2 x}$

Recall that $\frac{1}{\cos x} = \sec x$. Therefore, $\frac{1}{\cos^2 x} = \sec^2 x$.

The value of the limit is $\sec^2 x$. This is the derivative of $\tan x$.

Given limit:

$\lim\limits_{x \to -1} \frac{x^3 + 1}{x + 1}$

As $x \to -1$, the numerator $x^3 + 1 \to (-1)^3 + 1 = -1 + 1 = 0$.

As $x \to -1$, the denominator $x + 1 \to -1 + 1 = 0$.

The limit is of the indeterminate form $\frac{0}{0}$.

Method 1: Factoring

Factor the numerator using the sum of cubes formula, $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$.

Here, $a = x$ and $b = 1$.

$x^3 + 1^3 = (x+1)(x^2 - x(1) + 1^2) = (x+1)(x^2 - x + 1)$.

Substitute the factored form into the limit expression:

$\lim\limits_{x \to -1} \frac{(x+1)(x^2 - x + 1)}{x + 1}$

Since $x \to -1$, $x \neq -1$, which means $x+1 \neq 0$. We can cancel the common factor $(x+1)$.

$\lim\limits_{x \to -1} \frac{\cancel{(x+1)}(x^2 - x + 1)}{\cancel{(x+1)}} = \lim\limits_{x \to -1} (x^2 - x + 1)$

Now, substitute $x = -1$ into the simplified expression, as it is a polynomial and continuous everywhere:

$(-1)^2 - (-1) + 1 = 1 + 1 + 1 = 3$

Thus, $\lim\limits_{x \to -1} \frac{x^3 + 1}{x + 1} = 3$.

Alternate Solution: Using L'Hopital's Rule

Since the limit is of the indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule.

Let $f(x) = x^3 + 1$ and $g(x) = x + 1$.

Find the derivative of the numerator $f(x)$:

$f'(x) = \frac{d}{dx}(x^3 + 1) = 3x^2$

Find the derivative of the denominator $g(x)$:

$g'(x) = \frac{d}{dx}(x + 1) = 1$

Apply L'Hopital's Rule:

$\lim\limits_{x \to -1} \frac{x^3 + 1}{x + 1} = \lim\limits_{x \to -1} \frac{f'(x)}{g'(x)} = \lim\limits_{x \to -1} \frac{3x^2}{1}$

$= \lim\limits_{x \to -1} 3x^2$

Substitute $x=-1$ into the expression:

$3(-1)^2 = 3(1) = 3$

Both methods yield the same result.

The final answer is 3.

Given limit:

$\lim\limits_{x \to 0} \frac{x \sec x}{5x}$

As $x \to 0$, the expression is $\frac{0 \cdot \sec 0}{5 \cdot 0} = \frac{0 \cdot 1}{0} = \frac{0}{0}$, which is an indeterminate form.

We can simplify the expression before evaluating the limit, provided $x \neq 0$. Since we are considering the limit as $x \to 0$, we are looking at values of $x$ close to 0 but not equal to 0.

Cancel the common factor $x$ in the numerator and the denominator (for $x \neq 0$):

$\frac{x \sec x}{5x} = \frac{\cancel{x} \sec x}{5\cancel{x}} = \frac{\sec x}{5}$

Now, evaluate the limit of the simplified expression:

$\lim\limits_{x \to 0} \frac{\sec x}{5}$

Since $\sec x = \frac{1}{\cos x}$, we have:

$\lim\limits_{x \to 0} \frac{\frac{1}{\cos x}}{5} = \lim\limits_{x \to 0} \frac{1}{5 \cos x}$

As $x \to 0$, $\cos x \to \cos 0 = 1$ (since $\cos x$ is continuous at $x=0$).

Substitute the limit value of $\cos x$:

$\lim\limits_{x \to 0} \frac{1}{5 \cos x} = \frac{1}{5 \cdot 1} = \frac{1}{5}$

The final answer is $\frac{1}{5}$.

Given function:

$f(x) = (x-a)(x-b)$

We need to find the derivative of $f(x)$. We can use the Product Rule or by expanding the expression first.

Method 1: Using the Product Rule

The Product Rule states that if $f(x) = u(x)v(x)$, then the derivative is $f'(x) = u'(x)v(x) + u(x)v'(x)$.

Let $u(x) = x-a$ and $v(x) = x-b$.

Find the derivatives of $u(x)$ and $v(x)$ with respect to $x$:

$u'(x) = \frac{d}{dx}(x-a)$

$u'(x) = \frac{d}{dx}(x) - \frac{d}{dx}(a)$

$u'(x) = 1 - 0 = 1$

$v'(x) = \frac{d}{dx}(x-b)$

$v'(x) = \frac{d}{dx}(x) - \frac{d}{dx}(b)$

$v'(x) = 1 - 0 = 1$

Apply the Product Rule formula $f'(x) = u'(x)v(x) + u(x)v'(x)$:

$f'(x) = (1)(x-b) + (x-a)(1)$

Simplify the expression:

$f'(x) = x - b + x - a$

$f'(x) = 2x - a - b$

$f'(x) = 2x - (a+b)$

Method 2: Expanding the expression

Expand the given function by multiplying the terms:

$f(x) = (x-a)(x-b)$

$f(x) = x \cdot x - x \cdot b - a \cdot x + a \cdot b$

$f(x) = x^2 - bx - ax + ab$

$f(x) = x^2 - (a+b)x + ab$

Now, differentiate the expanded polynomial term by term using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ and the constant rule $\frac{d}{dx}(c) = 0$:

$f'(x) = \frac{d}{dx}(x^2 - (a+b)x + ab)$

$f'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}((a+b)x) + \frac{d}{dx}(ab)$

Note that $(a+b)$ and $ab$ are constants since $a$ and $b$ are constants.

$f'(x) = 2x^{2-1} - (a+b) \frac{d}{dx}(x) + 0$

$f'(x) = 2x - (a+b) \cdot 1$

$f'(x) = 2x - (a+b)$

Both methods yield the same result.

The derivative of $f(x) = (x-a)(x-b)$ is $f'(x) = 2x - (a+b)$.

Given limit:

$\lim\limits_{x \to a} \frac{(x+2)^{5/3} - (a+2)^{5/3}}{x - a}$

As $x \to a$, the numerator $(x+2)^{5/3} - (a+2)^{5/3} \to (a+2)^{5/3} - (a+2)^{5/3} = 0$.

As $x \to a$, the denominator $x - a \to a - a = 0$.

The limit is of the indeterminate form $\frac{0}{0}$.

Method 1: Using the Definition of the Derivative

Recall the definition of the derivative of a function $f(x)$ at a point $x=a$:

$f'(a) = \lim\limits_{x \to a} \frac{f(x) - f(a)}{x - a}$

Comparing the given limit expression with this definition, we can identify the function $f(x)$ and the point $a$.

Let $f(x) = (x+2)^{5/3}$.

Then $f(a) = (a+2)^{5/3}$.

The given limit is exactly the derivative of $f(x) = (x+2)^{5/3}$ evaluated at $x=a$.

We need to find the derivative of $f(x) = (x+2)^{5/3}$. We use the chain rule.

Let $u = x+2$. Then $f(x) = u^{5/3}$.

The chain rule states $\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx}$.

Find $\frac{df}{du}$ using the power rule $\frac{d}{du}(u^n) = nu^{n-1}$:

$\frac{df}{du} = \frac{d}{du}(u^{5/3}) = \frac{5}{3} u^{(5/3 - 1)} = \frac{5}{3} u^{2/3}$

Find $\frac{du}{dx}$:

$\frac{du}{dx} = \frac{d}{dx}(x+2) = \frac{d}{dx}(x) + \frac{d}{dx}(2) = 1 + 0 = 1$

Apply the chain rule:

$f'(x) = \frac{df}{du} \cdot \frac{du}{dx} = \frac{5}{3} u^{2/3} \cdot 1 = \frac{5}{3} (x+2)^{2/3}$

Now, evaluate $f'(x)$ at $x=a$:

$f'(a) = \frac{5}{3} (a+2)^{2/3}$

Therefore, the value of the limit is $f'(a)$.

$\lim\limits_{x \to a} \frac{(x+2)^{5/3} - (a+2)^{5/3}}{x - a} = \frac{5}{3} (a+2)^{2/3}$

Alternate Solution: Using L'Hopital's Rule

Since the limit is of the indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule.

L'Hopital's Rule states that if $\lim\limits_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim\limits_{x \to c} \frac{f(x)}{g(x)} = \lim\limits_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.

Let $N(x) = (x+2)^{5/3} - (a+2)^{5/3}$ (Numerator) and $D(x) = x - a$ (Denominator).

Find the derivative of the numerator with respect to $x$. Note that $(a+2)^{5/3}$ is a constant with respect to $x$, so its derivative is 0.

$N'(x) = \frac{d}{dx} \left( (x+2)^{5/3} - (a+2)^{5/3} \right)$

$N'(x) = \frac{d}{dx} (x+2)^{5/3} - \frac{d}{dx} (a+2)^{5/3}$

Using the chain rule for $(x+2)^{5/3}$: $\frac{d}{dx}((x+2)^{5/3}) = \frac{5}{3}(x+2)^{5/3 - 1} \cdot \frac{d}{dx}(x+2) = \frac{5}{3}(x+2)^{2/3} \cdot 1 = \frac{5}{3}(x+2)^{2/3}$.

$N'(x) = \frac{5}{3}(x+2)^{2/3} - 0 = \frac{5}{3}(x+2)^{2/3}$.

Find the derivative of the denominator with respect to $x$:

$D'(x) = \frac{d}{dx} (x - a) = \frac{d}{dx}(x) - \frac{d}{dx}(a) = 1 - 0 = 1$.

Apply L'Hopital's Rule:

$\lim\limits_{x \to a} \frac{(x+2)^{5/3} - (a+2)^{5/3}}{x - a} = \lim\limits_{x \to a} \frac{N'(x)}{D'(x)}$

$= \lim\limits_{x \to a} \frac{\frac{5}{3}(x+2)^{2/3}}{1}$

$= \lim\limits_{x \to a} \frac{5}{3}(x+2)^{2/3}$

Now, substitute $x=a$ into the expression (since it is a continuous function):

$= \frac{5}{3}(a+2)^{2/3}$

Both methods yield the same result.

The final answer is $\frac{5}{3} (a+2)^{2/3}$.

Given limit:

$\lim\limits_{x \to 0} \frac{\sin ax \cdot \cos bx}{\sin cx}$

As $x \to 0$, the numerator approaches $\sin(a \cdot 0) \cdot \cos(b \cdot 0) = \sin 0 \cdot \cos 0 = 0 \cdot 1 = 0$.

As $x \to 0$, the denominator approaches $\sin(c \cdot 0) = \sin 0 = 0$.

The limit is of the indeterminate form $\frac{0}{0}$.

Method 1: Using Standard Limits

We will use the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

Rewrite the expression by dividing the numerator and the denominator by $x$:

$\frac{\sin ax \cdot \cos bx}{\sin cx} = \frac{\frac{\sin ax \cdot \cos bx}{x}}{\frac{\sin cx}{x}}$

$= \frac{\frac{\sin ax}{x} \cdot \cos bx}{\frac{\sin cx}{x}}$

Now, manipulate the terms involving $\sin$ to match the standard limit form. Multiply the numerator and denominator of $\frac{\sin ax}{x}$ by $a$, and multiply the numerator and denominator of $\frac{\sin cx}{x}$ by $c$.

$= \frac{\frac{\sin ax}{ax} \cdot a \cdot \cos bx}{\frac{\sin cx}{cx} \cdot c}$

$= \frac{a}{c} \cdot \frac{\frac{\sin ax}{ax} \cdot \cos bx}{\frac{\sin cx}{cx}}$

Now, take the limit as $x \to 0$:

$\lim\limits_{x \to 0} \frac{a}{c} \cdot \frac{\frac{\sin ax}{ax} \cdot \cos bx}{\frac{\sin cx}{cx}}$

Using the properties of limits (limit of a product/quotient is the product/quotient of limits):

$= \frac{a}{c} \cdot \frac{\lim\limits_{x \to 0} \left(\frac{\sin ax}{ax}\right) \cdot \lim\limits_{x \to 0} (\cos bx)}{\lim\limits_{x \to 0} \left(\frac{\sin cx}{cx}\right)}$

Evaluate the individual limits:

For $\lim\limits_{x \to 0} \frac{\sin ax}{ax}$, let $\theta = ax$. As $x \to 0$, $\theta \to 0$. So, $\lim\limits_{x \to 0} \frac{\sin ax}{ax} = \lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

For $\lim\limits_{x \to 0} \cos bx$, since $\cos x$ is continuous, we can substitute $x=0$: $\cos(b \cdot 0) = \cos 0 = 1$.

For $\lim\limits_{x \to 0} \frac{\sin cx}{cx}$, let $\phi = cx$. As $x \to 0$, $\phi \to 0$. So, $\lim\limits_{x \to 0} \frac{\sin cx}{cx} = \lim\limits_{\phi \to 0} \frac{\sin \phi}{\phi} = 1$.

Substitute these values back into the expression:

$= \frac{a}{c} \cdot \frac{1 \cdot 1}{1}$

$= \frac{a}{c}$

Alternate Solution: Using L'Hopital's Rule

Since the limit is of the indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule.

L'Hopital's Rule states that if $\lim\limits_{x \to p} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim\limits_{x \to p} \frac{f(x)}{g(x)} = \lim\limits_{x \to p} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.

Let $f(x) = \sin ax \cdot \cos bx$ and $g(x) = \sin cx$.

Find the derivative of $f(x)$ using the product rule and chain rule:

$f'(x) = \frac{d}{dx}(\sin ax \cdot \cos bx)$

$= \frac{d}{dx}(\sin ax) \cdot \cos bx + \sin ax \cdot \frac{d}{dx}(\cos bx)$

$= (a \cos ax) \cdot \cos bx + \sin ax \cdot (-b \sin bx)$

$= a \cos ax \cos bx - b \sin ax \sin bx$

Find the derivative of $g(x)$ using the chain rule:

$g'(x) = \frac{d}{dx}(\sin cx)$

$= c \cos cx$

Apply L'Hopital's Rule:

$\lim\limits_{x \to 0} \frac{\sin ax \cdot \cos bx}{\sin cx} = \lim\limits_{x \to 0} \frac{a \cos ax \cos bx - b \sin ax \sin bx}{c \cos cx}$

Now, substitute $x=0$ into the expression (since the denominator is non-zero at $x=0$ as $c \neq 0$ and $\cos 0 = 1$):

$= \frac{a \cos(a \cdot 0) \cos(b \cdot 0) - b \sin(a \cdot 0) \sin(b \cdot 0)}{c \cos(c \cdot 0)}$

$= \frac{a \cos 0 \cos 0 - b \sin 0 \sin 0}{c \cos 0}$

$= \frac{a(1)(1) - b(0)(0)}{c(1)}$

$= \frac{a - 0}{c} = \frac{a}{c}$

Both methods yield the same result.

The final answer is $\frac{a}{c}$.

Given function:

$f(x) = \sqrt{x}$

We need to find the derivative of $f(x) = \sqrt{x}$ using the first principle, which is given by the formula:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

First, find $f(x+h)$:

$f(x+h) = \sqrt{x+h}$

Now, find the difference $f(x+h) - f(x)$:

$f(x+h) - f(x) = \sqrt{x+h} - \sqrt{x}$

Form the difference quotient $\frac{f(x+h) - f(x)}{h}$:

$\frac{f(x+h) - f(x)}{h} = \frac{\sqrt{x+h} - \sqrt{x}}{h}$

To evaluate the limit as $h \to 0$, which is of the form $\frac{0}{0}$, we multiply the numerator and the denominator by the conjugate of the numerator, which is $\sqrt{x+h} + \sqrt{x}$.

$f'(x) = \lim\limits_{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})}$

Use the difference of squares formula $(a-b)(a+b) = a^2 - b^2$ in the numerator:

$(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x}) = (\sqrt{x+h})^2 - (\sqrt{x})^2 = (x+h) - x = h$

Substitute this back into the expression:

$f'(x) = \lim\limits_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}$

For $h \neq 0$, we can cancel the term $h$ from the numerator and the denominator:

$f'(x) = \lim\limits_{h \to 0} \frac{\cancel{h}}{\cancel{h}(\sqrt{x+h} + \sqrt{x})}$

$f'(x) = \lim\limits_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}$

Now, evaluate the limit by substituting $h=0$ (since the denominator is not zero for $x > 0$):

$f'(x) = \frac{1}{\sqrt{x+0} + \sqrt{x}}$

$f'(x) = \frac{1}{\sqrt{x} + \sqrt{x}}$

$f'(x) = \frac{1}{2\sqrt{x}}$

The derivative of $f(x) = \sqrt{x}$ is $f'(x) = \frac{1}{2\sqrt{x}}$.

Given limit:

$\lim\limits_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1}$

As $x \to 0$, the numerator $\cos 2x - 1 \to \cos(2 \cdot 0) - 1 = \cos 0 - 1 = 1 - 1 = 0$.

As $x \to 0$, the denominator $\cos x - 1 \to \cos 0 - 1 = 1 - 1 = 0$.

The limit is of the indeterminate form $\frac{0}{0}$.

Method 1: Using Trigonometric Identities and Standard Limits

Recall the double angle identity: $\cos 2x = 1 - 2 \sin^2 x$.

So, $\cos 2x - 1 = (1 - 2 \sin^2 x) - 1 = -2 \sin^2 x$.

Recall the half-angle identity: $\cos x = 1 - 2 \sin^2 \left(\frac{x}{2}\right)$.

So, $\cos x - 1 = (1 - 2 \sin^2 \left(\frac{x}{2}\right)) - 1 = -2 \sin^2 \left(\frac{x}{2}\right)$.

Substitute these into the limit expression:

$\lim\limits_{x \to 0} \frac{-2 \sin^2 x}{-2 \sin^2 \left(\frac{x}{2}\right)} = \lim\limits_{x \to 0} \frac{\sin^2 x}{\sin^2 \left(\frac{x}{2}\right)}$

$= \lim\limits_{x \to 0} \left( \frac{\sin x}{\sin \left(\frac{x}{2}\right)} \right)^2$

We use the standard limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. Manipulate the expression inside the square:

$\frac{\sin x}{\sin \left(\frac{x}{2}\right)} = \frac{\frac{\sin x}{x} \cdot x}{\frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}} \cdot \frac{x}{2}}$

$= \frac{\frac{\sin x}{x}}{\frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}}} \cdot \frac{x}{\frac{x}{2}}$

$= \frac{\frac{\sin x}{x}}{\frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}}} \cdot 2$

Now, evaluate the limit of this expression as $x \to 0$:

$\lim\limits_{x \to 0} \left( \frac{\frac{\sin x}{x}}{\frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}}} \cdot 2 \right)$

$= \frac{\lim\limits_{x \to 0} \frac{\sin x}{x}}{\lim\limits_{x \to 0} \frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}}} \cdot 2$

As $x \to 0$, $x \to 0$ and $\frac{x}{2} \to 0$. Using the standard limit:

$\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$

$\lim\limits_{x \to 0} \frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}} = 1$

Substitute these values:

$= \frac{1}{1} \cdot 2 = 2$

Now, square the result to get the original limit:

$\lim\limits_{x \to 0} \left( \frac{\sin x}{\sin \left(\frac{x}{2}\right)} \right)^2 = (2)^2 = 4$

Alternate Solution: Using L'Hopital's Rule

Since the limit is of the indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule.

Let $f(x) = \cos 2x - 1$ and $g(x) = \cos x - 1$.

Find the derivative of $f(x)$:

$f'(x) = \frac{d}{dx}(\cos 2x - 1) = - \sin(2x) \cdot \frac{d}{dx}(2x) - 0 = -2 \sin 2x$

Find the derivative of $g(x)$:

$g'(x) = \frac{d}{dx}(\cos x - 1) = - \sin x - 0 = -\sin x$

Apply L'Hopital's Rule for the first time:

$\lim\limits_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1} = \lim\limits_{x \to 0} \frac{-2 \sin 2x}{-\sin x} = \lim\limits_{x \to 0} \frac{2 \sin 2x}{\sin x}$

This new limit is also of the indeterminate form $\frac{0}{0}$. Apply L'Hopital's Rule again.

Find the derivative of the new numerator $2 \sin 2x$:

$\frac{d}{dx}(2 \sin 2x) = 2 \cos(2x) \cdot \frac{d}{dx}(2x) = 2 \cos(2x) \cdot 2 = 4 \cos 2x$

Find the derivative of the new denominator $\sin x$:

$\frac{d}{dx}(\sin x) = \cos x$

Apply L'Hopital's Rule for the second time:

$\lim\limits_{x \to 0} \frac{2 \sin 2x}{\sin x} = \lim\limits_{x \to 0} \frac{4 \cos 2x}{\cos x}$

Now, substitute $x=0$ into the expression (since the denominator is non-zero at $x=0$):

$= \frac{4 \cos(2 \cdot 0)}{\cos 0} = \frac{4 \cos 0}{\cos 0} = \frac{4 \cdot 1}{1} = 4$

Both methods yield the same result.

The final answer is 4.

Given function:

$f(x) = \tan x$

We need to find the derivative of $f(x)$ using the Quotient Rule.

Recall that $\tan x = \frac{\sin x}{\cos x}$.

The Quotient Rule states that if $f(x) = \frac{u(x)}{v(x)}$, then the derivative $f'(x)$ is given by:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

Let $u(x) = \sin x$ and $v(x) = \cos x$.

We are given the derivatives of $\sin x$ and $\cos x$:

$u'(x) = \frac{d}{dx}(\sin x) = \cos x$

$v'(x) = \frac{d}{dx}(\cos x) = -\sin x$

Now, apply the Quotient Rule formula $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$:

$f'(x) = \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{(\cos x)^2}$

Simplify the numerator:

$(\cos x)(\cos x) - (\sin x)(-\sin x) = \cos^2 x + \sin^2 x$

Substitute this back into the expression for $f'(x)$:

$f'(x) = \frac{\cos^2 x + \sin^2 x}{\cos^2 x}$

Using the fundamental trigonometric identity $\sin^2 x + \cos^2 x = 1$:

$f'(x) = \frac{1}{\cos^2 x}$

Recall that $\sec x = \frac{1}{\cos x}$. So, $\frac{1}{\cos^2 x} = \sec^2 x$.

$f'(x) = \sec^2 x$

The derivative of $f(x) = \tan x$ is $f'(x) = \sec^2 x$.

Given limit:

$\lim\limits_{x \to 0} \frac{e^x - e^{-x}}{x}$

As $x \to 0$, the numerator $e^x - e^{-x} \to e^0 - e^{-0} = 1 - 1 = 0$.

As $x \to 0$, the denominator $x \to 0$.

The limit is of the indeterminate form $\frac{0}{0}$.

Method 1: Using L'Hopital's Rule

Since the limit is of the indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule.

L'Hopital's Rule states that if $\lim\limits_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim\limits_{x \to c} \frac{f(x)}{g(x)} = \lim\limits_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.

Let $f(x) = e^x - e^{-x}$ and $g(x) = x$.

Find the derivative of the numerator $f(x)$:

$f'(x) = \frac{d}{dx}(e^x - e^{-x})$

$= \frac{d}{dx}(e^x) - \frac{d}{dx}(e^{-x})$

$= e^x - (e^{-x} \cdot \frac{d}{dx}(-x))$ (using chain rule for $e^{-x}$)

$= e^x - (e^{-x} \cdot (-1))$

$f'(x) = e^x + e^{-x}$

Find the derivative of the denominator $g(x)$:

$g'(x) = \frac{d}{dx}(x)$

$g'(x) = 1$

Apply L'Hopital's Rule:

$\lim\limits_{x \to 0} \frac{e^x - e^{-x}}{x} = \lim\limits_{x \to 0} \frac{f'(x)}{g'(x)} = \lim\limits_{x \to 0} \frac{e^x + e^{-x}}{1}$

Now, substitute $x = 0$ into the new expression (since the denominator is not zero at $x=0$):

$= \frac{e^0 + e^{-0}}{1}$

$= \frac{1 + 1}{1}$

$= \frac{2}{1} = 2$

Alternate Solution: Using Standard Limit

We use the standard limit $\lim\limits_{y \to 0} \frac{e^y - 1}{y} = 1$.

Rewrite the given expression by splitting the numerator:

$\frac{e^x - e^{-x}}{x} = \frac{e^x - 1 + 1 - e^{-x}}{x}$

$= \frac{(e^x - 1) - (e^{-x} - 1)}{x}$

$= \frac{e^x - 1}{x} - \frac{e^{-x} - 1}{x}$

Now, evaluate the limit of each term:

$\lim\limits_{x \to 0} \left( \frac{e^x - 1}{x} - \frac{e^{-x} - 1}{x} \right) = \lim\limits_{x \to 0} \frac{e^x - 1}{x} - \lim\limits_{x \to 0} \frac{e^{-x} - 1}{x}$

The first limit is the standard limit:

$\lim\limits_{x \to 0} \frac{e^x - 1}{x} = 1$

For the second limit, let $y = -x$. As $x \to 0$, $y \to 0$. Also, $x = -y$.

$\lim\limits_{x \to 0} \frac{e^{-x} - 1}{x} = \lim\limits_{y \to 0} \frac{e^y - 1}{-y}$

$= -\lim\limits_{y \to 0} \frac{e^y - 1}{y}$

$= -1$ (using the standard limit)

Substitute these values back into the original limit expression:

$\lim\limits_{x \to 0} \frac{e^x - e^{-x}}{x} = 1 - (-1) = 1 + 1 = 2$

Both methods yield the same result.

The final answer is 2.

Given:

The function is $f(x) = \frac{x}{\sin x}$.

Solution:

We need to find the derivative of $f(x) = \frac{x}{\sin x}$ using the Quotient Rule.

The Quotient Rule for differentiation states that if $f(x) = \frac{u(x)}{v(x)}$, where $u(x)$ and $v(x)$ are differentiable functions, then the derivative of $f(x)$ is given by:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

In our function $f(x) = \frac{x}{\sin x}$, we identify the numerator and the denominator:

Let $u(x) = x$ (the numerator).

Let $v(x) = \sin x$ (the denominator).

Now, find the derivatives of $u(x)$ and $v(x)$ with respect to $x$:

The derivative of $u(x) = x$ is:

$u'(x) = \frac{d}{dx}(x) = 1$

The derivative of $v(x) = \sin x$ is:

$v'(x) = \frac{d}{dx}(\sin x) = \cos x$

Substitute $u(x)$, $v(x)$, $u'(x)$, and $v'(x)$ into the Quotient Rule formula:

$f'(x) = \frac{(1)(\sin x) - (x)(\cos x)}{(\sin x)^2}$

Simplify the expression:

$f'(x) = \frac{\sin x - x \cos x}{\sin^2 x}$

The derivative of $f(x) = \frac{x}{\sin x}$ is $f'(x) = \frac{\sin x - x \cos x}{\sin^2 x}$.

Given limit:

$\lim\limits_{x \to 0} \frac{\sin(a+x) - \sin(a-x)}{x}$

As $x \to 0$, the numerator approaches $\sin(a+0) - \sin(a-0) = \sin a - \sin a = 0$.

As $x \to 0$, the denominator approaches $0$.

The limit is of the indeterminate form $\frac{0}{0}$.

Method 1: Using Trigonometric Identities and Standard Limits

We use the sum-to-product trigonometric identity for the numerator:

$\sin A - \sin B = 2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$

Let $A = a+x$ and $B = a-x$.

Then $\frac{A+B}{2} = \frac{(a+x) + (a-x)}{2} = \frac{2a}{2} = a$.

And $\frac{A-B}{2} = \frac{(a+x) - (a-x)}{2} = \frac{2x}{2} = x$.

Substitute these into the sum-to-product formula:

$\sin(a+x) - \sin(a-x) = 2 \cos(a) \sin(x)$

Substitute this back into the limit expression:

$\lim\limits_{x \to 0} \frac{2 \cos a \sin x}{x}$

We can separate the constant term $2 \cos a$ from the limit:

$= 2 \cos a \cdot \lim\limits_{x \to 0} \frac{\sin x}{x}$

We know the standard limit $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$.

Substitute the value of the standard limit:

$= 2 \cos a \cdot 1$

$= 2 \cos a$

Alternate Solution: Using L'Hopital's Rule

Since the limit is of the indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule.

Let $f(x) = \sin(a+x) - \sin(a-x)$ and $g(x) = x$.

Find the derivative of the numerator $f(x)$ with respect to $x$ (remember $a$ is a constant):

$f'(x) = \frac{d}{dx}(\sin(a+x) - \sin(a-x))$

$= \frac{d}{dx}(\sin(a+x)) - \frac{d}{dx}(\sin(a-x))$

Using the chain rule: $\frac{d}{dx}(\sin(u)) = \cos(u) \cdot \frac{du}{dx}$.

For $\sin(a+x)$, let $u = a+x$, so $\frac{du}{dx} = 1$. $\frac{d}{dx}(\sin(a+x)) = \cos(a+x) \cdot 1 = \cos(a+x)$.

For $\sin(a-x)$, let $v = a-x$, so $\frac{dv}{dx} = -1$. $\frac{d}{dx}(\sin(a-x)) = \cos(a-x) \cdot (-1) = -\cos(a-x)$.

$f'(x) = \cos(a+x) - (-\cos(a-x))$

$f'(x) = \cos(a+x) + \cos(a-x)$

Find the derivative of the denominator $g(x)$ with respect to $x$:

$g'(x) = \frac{d}{dx}(x) = 1$

Apply L'Hopital's Rule:

$\lim\limits_{x \to 0} \frac{\sin(a+x) - \sin(a-x)}{x} = \lim\limits_{x \to 0} \frac{\cos(a+x) + \cos(a-x)}{1}$

Now, substitute $x = 0$ into the expression (since the denominator is not zero at $x=0$):

$= \cos(a+0) + \cos(a-0)$

$= \cos a + \cos a$

$= 2 \cos a$

Both methods yield the same result.

The final answer is $2 \cos a$.

Given function:

$f(x) = \sin(x^2 + 1)$

We need to find the derivative of $f(x)$ using the Chain Rule.

The Chain Rule states that if $y = f(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

In our function, we can identify the outer function and the inner function.

Let the outer function be $\sin(u)$ and the inner function be $u = x^2 + 1$.

So, $f(x) = \sin(u)$ where $u = x^2 + 1$.

Now, find the derivative of the outer function with respect to $u$:

$\frac{df}{du} = \frac{d}{du}(\sin u)$

We know the derivative of $\sin u$ with respect to $u$ is $\cos u$.

$\frac{df}{du} = \cos u$

Next, find the derivative of the inner function with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(x^2 + 1)$

Using the power rule and constant rule for differentiation:

$\frac{du}{dx} = \frac{d}{dx}(x^2) + \frac{d}{dx}(1)$

$\frac{du}{dx} = 2x + 0 = 2x$

Now, apply the Chain Rule $\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx}$:

$f'(x) = (\cos u) \cdot (2x)$

Substitute back $u = x^2 + 1$ into the expression:

$f'(x) = (\cos(x^2 + 1)) \cdot (2x)$

$f'(x) = 2x \cos(x^2 + 1)$

The derivative of $f(x) = \sin(x^2 + 1)$ is $f'(x) = 2x \cos(x^2 + 1)$.

Given limit:

$\lim\limits_{x \to 0} \frac{\log_e(1+x)}{x}$

As $x \to 0$, the numerator $\log_e(1+x) \to \log_e(1+0) = \log_e(1) = 0$.

As $x \to 0$, the denominator $x \to 0$.

The limit is of the indeterminate form $\frac{0}{0}$.

Method 1: Using Standard Limit

We know the standard limit:

$\lim\limits_{x \to 0} \frac{\log_e(1+x)}{x} = 1$

Using this standard result, the evaluation is direct.

$\lim\limits_{x \to 0} \frac{\log_e(1+x)}{x} = 1$

Method 2: Using L'Hopital's Rule

Since the limit is of the indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule.

L'Hopital's Rule states that if $\lim\limits_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim\limits_{x \to c} \frac{f(x)}{g(x)} = \lim\limits_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.

Let $f(x) = \log_e(1+x)$ and $g(x) = x$.

Find the derivative of the numerator $f(x)$:

$f'(x) = \frac{d}{dx}(\log_e(1+x)) = \frac{1}{1+x} \cdot \frac{d}{dx}(1+x)$

$f'(x) = \frac{1}{1+x} \cdot 1 = \frac{1}{1+x}$

Find the derivative of the denominator $g(x)$:

$g'(x) = \frac{d}{dx}(x)$

$g'(x) = 1$

Apply L'Hopital's Rule:

$\lim\limits_{x \to 0} \frac{\log_e(1+x)}{x} = \lim\limits_{x \to 0} \frac{f'(x)}{g'(x)} = \lim\limits_{x \to 0} \frac{\frac{1}{1+x}}{1}$

$= \lim\limits_{x \to 0} \frac{1}{1+x}$

Now, substitute $x = 0$ into the expression (since the denominator is not zero at $x=0$):

$= \frac{1}{1+0} = \frac{1}{1} = 1$

Both methods yield the same result.

The final answer is 1.

Given:

The function is $f(x) = \cos x \cdot \text{cot } x$.

Solution:

We need to find the derivative of $f(x) = \cos x \cdot \text{cot } x$ using the Product Rule.

The Product Rule for differentiation states that if $f(x) = u(x)v(x)$, where $u(x)$ and $v(x)$ are differentiable functions, then the derivative of $f(x)$ is given by:

$f'(x) = u'(x)v(x) + u(x)v'(x)$

In our function $f(x) = \cos x \cdot \text{cot } x$, we identify $u(x)$ and $v(x)$:

Let $u(x) = \cos x$.

Let $v(x) = \text{cot } x$.

Now, find the derivatives of $u(x)$ and $v(x)$ with respect to $x$:

The derivative of $u(x) = \cos x$ is:

$u'(x) = \frac{d}{dx}(\cos x) = -\sin x$

The derivative of $v(x) = \text{cot } x$ is:

$v'(x) = \frac{d}{dx}(\text{cot } x) = -\text{cosec}^2 x$

Substitute $u(x)$, $v(x)$, $u'(x)$, and $v'(x)$ into the Product Rule formula $f'(x) = u'(x)v(x) + u(x)v'(x)$:

$f'(x) = (-\sin x)(\text{cot } x) + (\cos x)(-\text{cosec}^2 x)$

Simplify the expression. Recall that $\text{cot } x = \frac{\cos x}{\sin x}$ and $\text{cosec}^2 x = \frac{1}{\sin^2 x}$.

$f'(x) = (-\sin x) \left(\frac{\cos x}{\sin x}\right) + (\cos x) \left(-\frac{1}{\sin^2 x}\right)$

$f'(x) = -\cos x - \frac{\cos x}{\sin^2 x}$

The derivative can also be written as:

$f'(x) = -\cos x - \cos x \cdot \frac{1}{\sin^2 x}$

$f'(x) = -\cos x (1 + \frac{1}{\sin^2 x})$

$f'(x) = -\cos x (1 + \text{cosec}^2 x)$

The derivative of $f(x) = \cos x \cdot \text{cot } x$ is $f'(x) = -\cos x - \frac{\cos x}{\sin^2 x}$ or $f'(x) = -\cos x (1 + \text{cosec}^2 x)$.

Given limit:

$\lim\limits_{x \to \pi/2} \frac{\tan 2x}{x - \pi/2}$

Check the form of the limit:

As $x \to \pi/2$, the numerator approaches $\tan(2 \cdot \pi/2) = \tan(\pi) = 0$.

As $x \to \pi/2$, the denominator approaches $\pi/2 - \pi/2 = 0$.

The limit is of the indeterminate form $\frac{0}{0}$.

Method 1: Using Substitution and Standard Limits

Let $y = x - \pi/2$. As $x \to \pi/2$, $y \to 0$.

From the substitution, we have $x = y + \pi/2$.

Substitute $x$ in the numerator term $\tan 2x$:

$\tan(2x) = \tan(2(y + \pi/2)) = \tan(2y + \pi)$

Using the trigonometric property that the tangent function has a period of $\pi$, $\tan(\theta + \pi) = \tan(\theta)$.

So, $\tan(2y + \pi) = \tan(2y)$.

The limit expression in terms of $y$ becomes:

$\lim\limits_{y \to 0} \frac{\tan(2y)}{y}$

We need to evaluate this limit. We use the standard limit $\lim\limits_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$.

Manipulate the expression to match the standard limit form:

$\frac{\tan(2y)}{y} = \frac{\tan(2y)}{2y} \cdot 2$

Now, evaluate the limit:

$\lim\limits_{y \to 0} \left( \frac{\tan(2y)}{2y} \cdot 2 \right)$

Using the property of limits $\lim\limits_{y \to 0} (c \cdot f(y)) = c \cdot \lim\limits_{y \to 0} f(y)$:

$= 2 \cdot \lim\limits_{y \to 0} \frac{\tan(2y)}{2y}$

Let $\theta = 2y$. As $y \to 0$, $\theta \to 0$. The limit becomes:

$= 2 \cdot \lim\limits_{\theta \to 0} \frac{\tan \theta}{\theta}$

Using the standard limit $\lim\limits_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$:

$= 2 \cdot 1$

$= 2$

Alternate Solution: Using L'Hopital's Rule

Since the limit is of the indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule.

L'Hopital's Rule states that if $\lim\limits_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim\limits_{x \to c} \frac{f(x)}{g(x)} = \lim\limits_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.

Let $f(x) = \tan 2x$ (the numerator) and $g(x) = x - \pi/2$ (the denominator).

Find the derivative of the numerator $f(x)$ with respect to $x$ using the chain rule:

$f'(x) = \frac{d}{dx}(\tan 2x)$

$f'(x) = \sec^2(2x) \cdot \frac{d}{dx}(2x)$

$f'(x) = \sec^2(2x) \cdot 2 = 2 \sec^2 2x$

Find the derivative of the denominator $g(x)$ with respect to $x$:

$g'(x) = \frac{d}{dx}(x - \pi/2)$

$g'(x) = \frac{d}{dx}(x) - \frac{d}{dx}(\pi/2)$

$g'(x) = 1 - 0 = 1$

Apply L'Hopital's Rule by taking the limit of the ratio of the derivatives:

$\lim\limits_{x \to \pi/2} \frac{\tan 2x}{x - \pi/2} = \lim\limits_{x \to \pi/2} \frac{2 \sec^2 2x}{1}$

$= \lim\limits_{x \to \pi/2} 2 \sec^2 2x$

Now, evaluate the limit by substituting $x = \pi/2$ into the expression:

$= 2 \sec^2 (2 \cdot \pi/2)$

$= 2 \sec^2 \pi$

Recall that $\sec \pi = \frac{1}{\cos \pi}$. The value of $\cos \pi$ is $-1$.

So, $\sec \pi = \frac{1}{-1} = -1$.

Therefore, $\sec^2 \pi = (-1)^2 = 1$.

Substitute this value back into the limit:

$= 2 \cdot 1$

$= 2$

Both methods yield the same result.

The final answer is 2.

Given:

The function is $f(x) = (x^2 + 1) \cos x$.

Solution:

We need to find the derivative of $f(x) = (x^2 + 1) \cos x$. The function is a product of two functions, so we will use the Product Rule for differentiation.

The Product Rule states that if $f(x) = u(x)v(x)$, where $u(x)$ and $v(x)$ are differentiable functions, then the derivative of $f(x)$ is given by:

$f'(x) = u'(x)v(x) + u(x)v'(x)$

In our function $f(x) = (x^2 + 1) \cos x$, we identify the two functions:

Let $u(x) = x^2 + 1$.

Let $v(x) = \cos x$.

Now, find the derivatives of $u(x)$ and $v(x)$ with respect to $x$:

The derivative of $u(x) = x^2 + 1$ is:

$u'(x) = \frac{d}{dx}(x^2 + 1)$

$u'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(1)$

$u'(x) = 2x + 0$

$u'(x) = 2x$

The derivative of $v(x) = \cos x$ is:

$v'(x) = \frac{d}{dx}(\cos x)$

$v'(x) = -\sin x$

Substitute $u(x)$, $v(x)$, $u'(x)$, and $v'(x)$ into the Product Rule formula $f'(x) = u'(x)v(x) + u(x)v'(x)$:

$f'(x) = (2x)(\cos x) + (x^2 + 1)(-\sin x)$

Simplify the expression:

$f'(x) = 2x \cos x - (x^2 + 1) \sin x$

$f'(x) = 2x \cos x - x^2 \sin x - \sin x$

The derivative of $f(x) = (x^2 + 1) \cos x$ is $f'(x) = 2x \cos x - (x^2 + 1) \sin x$.