Question

The value of $\underset{x\to 0}{\lim }\left[{\int }_0^{x^2}\frac{{\sec }^2\left(t\right)}{x.\sin \left(x\right)}dt\right]$

• A. $3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

The correct option is C

$1$

Explanation of the correct option.

Compute the required value.

Given : $\underset{x\to 0}{\lim }\left[\frac{{\int }_0^{x^2}{\sec }^2\left(t\right)dt}{x.\sin \left(x\right)}\right]$

Using Leibnitz rule,

$=\underset{x\to 0}{\lim }\frac{\tan \left(x^2\right)-\tan \left(0\right)}{x\sin \left(x\right)}$

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

It can be written as,

$$\begin{gathered} =\underset{x\to 0}{\lim }\frac{\tan \left(x^2\right)}{x^2}\times x^2\times \frac{x}{\sin \left(x\right)}\times \frac{1}{x\times x} \\ =\underset{x\to 0}{\lim }\frac{\tan \left(x^2\right)}{x^2}\times \frac{x}{\sin \left(x\right)}\left[\underset{x\to 0}{\lim }\frac{\tan \left(x^2\right)}{x^2}=1,\underset{x\to 0}{\lim }\frac{\sin \left(x\right)}{x}=1\right] \\ =1\times 1 \\ =1 \end{gathered}$$

Therefore the value of $\underset{x\to 0}{\lim }\left[{\int }_0^{x^2}\frac{{\sec }^2\left(t\right)}{x.\sin \left(x\right)}dt\right]$ is $1$.

Hence, option $\left(C\right)$ is the correct option.