CBSE Class 11 Maths Exemplar Chapter 13: Limits and Derivatives NCERT Solutions

We are asked to evaluate the limit \lim_{x \to 3} \frac{x^2 - 9}{x - 3}.

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

Therefore, we need to simplify the expression first.

We can factor the numerator using the difference of squares formula, a^2 - b^2 = (a-b)(a+b). Here, x^2 - 9 = x^2 - 3^2 = (x-3)(x+3).

So, the expression becomes:

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

Since x \to 3, x \neq 3, we can cancel the (x-3) term from the numerator and the denominator:

= \lim_{x \to 3} (x+3)

Now, we can substitute x = 3 into the simplified expression:

= 3 + 3 = 6

Thus, the value of the limit is 6.

We need to evaluate the limit \lim_{x \to 1/2} \frac{4x^2 - 1}{2x - 1}.

Direct substitution of x = 1/2 yields \frac{4(1/2)^2 - 1}{2(1/2) - 1} = \frac{4(1/4) - 1}{1 - 1} = \frac{1 - 1}{0} = \frac{0}{0}, which is an indeterminate form.

We must simplify the expression.

The numerator 4x^2 - 1 can be written as (2x)^2 - 1^2, which is a difference of squares. It factors as (2x-1)(2x+1).

So, the limit becomes:

\lim_{x \to 1/2} \frac{4x^2 - 1}{2x - 1} = \lim_{x \to 1/2} \frac{(2x)^2 - 1^2}{2x - 1}

= \lim_{x \to 1/2} \frac{(2x+1)(2x-1)}{(2x-1)}

Since x \to 1/2, x \neq 1/2, which means 2x \neq 1, so 2x-1 \neq 0. We can cancel the (2x-1) term:

= \lim_{x \to 1/2} (2x+1)

Now, substitute x = 1/2 into the simplified expression:

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

The value of the limit is 2.

We need to evaluate the limit \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}.

Direct substitution of h = 0 results in \frac{\sqrt{x+0} - \sqrt{x}}{0} = \frac{\sqrt{x} - \sqrt{x}}{0} = \frac{0}{0}, an indeterminate form.

We can evaluate this limit using the standard limit formula \lim_{y \to a} \frac{y^n - a^n}{y - a} = na^{n-1}.

Let y = x+h. As h \to 0, y \to x. The expression can be rewritten as:

\lim_{h \to 0} \frac{(x+h)^{1/2} - x^{1/2}}{h}

To match the standard form, we can rewrite the denominator h as (x+h) - x:

= \lim_{h \to 0} \frac{(x+h)^{1/2} - x^{1/2}}{(x+h) - x}

Now, let y = x+h and a = x. As h \to 0, y \to x. The limit becomes:

= \lim_{y \to x} \frac{y^{1/2} - a^{1/2}}{y - a}

Using the standard formula with n = 1/2 and a = x:

= n a^{n-1} = \frac{1}{2} x^{1/2 - 1} = \frac{1}{2} x^{-1/2}

This can be written as:

= \frac{1}{2 x^{1/2}} = \frac{1}{2\sqrt{x}}

Thus, the value of the limit is \frac{1}{2\sqrt{x}}.

We need to evaluate \lim_{x \to 0} \frac{(x+2)^{1/3} - 2^{1/3}}{x}.

Direct substitution of x = 0 gives \frac{(0+2)^{1/3} - 2^{1/3}}{0} = \frac{2^{1/3} - 2^{1/3}}{0} = \frac{0}{0}, which is an indeterminate form.

We will use the standard limit formula \lim_{y \to a} \frac{y^n - a^n}{y - a} = na^{n-1}.

Let y = x+2. As x \to 0, y \to 2. The denominator x can be written as (x+2) - 2.

So the limit becomes:

\lim_{x \to 0} \frac{(x+2)^{1/3} - 2^{1/3}}{(x+2) - 2}

Now, let y = x+2 and a = 2. As x \to 0, y \to 2. The limit is in the form \lim_{y \to a} \frac{y^n - a^n}{y - a} with n = 1/3.

Applying the formula na^{n-1}:

= \frac{1}{3} \times 2^{1/3 - 1}

= \frac{1}{3} \times 2^{-2/3}

This can be written as:

= \frac{1}{3 \times 2^{2/3}}

The value of the limit is \frac{1}{3(2)^{2/3}}.

We need to evaluate \lim_{x \to 0} \frac{(1+x)^6 - 1}{(1+x)^2 - 1}.

Direct substitution of x = 0 gives \frac{(1+0)^6 - 1}{(1+0)^2 - 1} = \frac{1^6 - 1}{1^2 - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0}, an indeterminate form.

We can use the standard limit formula \lim_{y \to a} \frac{y^n - a^n}{y - a} = na^{n-1}.

Let y = 1+x. As x \to 0, y \to 1.

We can rewrite the expression by dividing the numerator and the denominator by x (or (1+x) - 1):

\lim_{x \to 0} \frac{(1+x)^6 - 1}{(1+x)^2 - 1} = \lim_{x \to 0} \frac{\frac{(1+x)^6 - 1}{x}}{\frac{(1+x)^2 - 1}{x}}

Now, we can apply the standard limit formula to the numerator and the denominator separately.

For the numerator, let y = 1+x and a = 1. As x \to 0, y \to 1. The numerator limit is \lim_{y \to 1} \frac{y^6 - 1^6}{y - 1}. Using the formula na^{n-1} with n=6 and a=1, this limit is 6 \times 1^{6-1} = 6.

For the denominator, let y = 1+x and a = 1. As x \to 0, y \to 1. The denominator limit is \lim_{y \to 1} \frac{y^2 - 1^2}{y - 1}. Using the formula na^{n-1} with n=2 and a=1, this limit is 2 \times 1^{2-1} = 2.

Therefore, the original limit is the ratio of these two limits:

= \frac{\lim_{x \to 0} \frac{(1+x)^6 - 1}{(1+x) - 1}}{\lim_{x \to 0} \frac{(1+x)^2 - 1}{(1+x) - 1}}

= \frac{6(1)^{6-1}}{2(1)^{2-1}}

= \frac{6 \times 1}{2 \times 1} = \frac{6}{2} = 3

The value of the limit is 3.

We need to evaluate the limit \lim_{x \to a} \frac{(2+x)^{5/2} - (a+2)^{5/2}}{x-a}.

Direct substitution of x = a gives \frac{(2+a)^{5/2} - (a+2)^{5/2}}{a-a} = \frac{0}{0}, which is an indeterminate form.

We will use the standard limit formula \lim_{y \to b} \frac{y^n - b^n}{y - b} = nb^{n-1}.

Let y = 2+x. As x \to a, y \to 2+a.

The denominator x-a can be written as (2+x) - (2+a).

So the limit becomes:

\lim_{x \to a} \frac{(2+x)^{5/2} - (2+a)^{5/2}}{(2+x) - (2+a)}

Now, let y = 2+x and b = 2+a. As x \to a, y \to b. The limit is in the form \lim_{y \to b} \frac{y^n - b^n}{y - b} with n = 5/2.

Applying the formula nb^{n-1}:

= \frac{5}{2} \times (2+a)^{5/2 - 1}

= \frac{5}{2} \times (2+a)^{3/2}

The value of the limit is \frac{5}{2}(2+a)^{3/2}.