2. Find:

(i) [tex]\lim _{x \rightarrow \frac{\pi}{2}^{-}} \frac{\tan ^2 x}{\sec ^3 x}[/tex]

(ii) [tex]\lim _{x \rightarrow 0^{+}} \frac{\sin x}{x + x^{3 / 2}}[/tex]

by using L'Hospital's Rule.



Answer :

Let's solve the given limits step-by-step using L'Hôpital's Rule.

### (i) [tex]\(\lim _{x \rightarrow \frac{\pi}{2}^{-}} \frac{\tan^2 x}{\sec^3 x}\)[/tex]

First, we will simplify the expression:

The tangent and secant functions can be rewritten in terms of sine and cosine:

[tex]\[ \tan x = \frac{\sin x}{\cos x} \quad \text{and} \quad \sec x = \frac{1}{\cos x} \][/tex]

So,

[tex]\[ \frac{\tan^2 x}{\sec^3 x} = \frac{\left( \frac{\sin x}{\cos x} \right)^2}{\left( \frac{1}{\cos x} \right)^3} = \frac{\frac{\sin^2 x}{\cos^2 x}}{\frac{1}{\cos^3 x}} = \frac{\sin^2 x}{\cos^2 x} \times \cos^3 x = \sin^2 x \cos x \][/tex]

Now, we need to find the limit:

[tex]\[ \lim_{x \rightarrow \frac{\pi}{2}^{-}} \sin^2 x \cos x \][/tex]

As [tex]\(x\)[/tex] approaches [tex]\(\frac{\pi}{2}\)[/tex] from the left, [tex]\(\sin x\)[/tex] approaches 1 and [tex]\(\cos x\)[/tex] approaches 0 from the positive side. Therefore,

[tex]\[ \sin^2 x \to 1 \quad \text{and} \quad \cos x \to 0 \][/tex]

Thus,

[tex]\[ \lim_{x \rightarrow \frac{\pi}{2}^{-}} \sin^2 x \cos x = 1 \cdot 0 = 0 \][/tex]

So the limit is:

[tex]\[ \boxed{0} \][/tex]

### (ii) [tex]\(\lim _{x \rightarrow 0^{+}} \frac{\sin x}{x+x^{3/2}}\)[/tex]

This is an indeterminate form of type [tex]\(\frac{0}{0}\)[/tex]. Therefore, we can apply L'Hôpital's Rule, which states that:

[tex]\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \quad \text{if the limit on the right-hand side exists} \][/tex]

Given:

[tex]\[ f(x) = \sin x \quad \text{and} \quad g(x) = x + x^{3/2} \][/tex]

Let's compute their derivatives:

[tex]\[ f'(x) = \cos x \][/tex]
[tex]\[ g'(x) = 1 + \frac{3}{2} x^{1/2} \][/tex]

Therefore, applying L'Hôpital's Rule:

[tex]\[ \lim_{x \rightarrow 0^{+}} \frac{\sin x}{x + x^{3/2}} = \lim_{x \rightarrow 0^{+}} \frac{\cos x}{1 + \frac{3}{2} x^{1/2}} \][/tex]

Now, let's evaluate this limit.

As [tex]\(x\)[/tex] approaches 0 from the right, [tex]\(\cos x\)[/tex] approaches 1 and [tex]\(\frac{3}{2} x^{1/2}\)[/tex] approaches 0. Thus,

[tex]\[ \lim_{x \rightarrow 0^{+}} \frac{1}{1 + 0} = 1 \][/tex]

Hence, the limit is:

[tex]\[ \boxed{1} \][/tex]

In conclusion:

(i) [tex]\(\lim _{x \rightarrow \frac{\pi}{2}^{-}} \frac{\tan^2 x}{\sec^3 x} = 0\)[/tex]

(ii) [tex]\(\lim _{x \rightarrow 0^{+}} \frac{\sin x}{x+x^{3 / 2}} = 1\)[/tex]