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]
