Question

The value of $\underset{x\to 0}{lim}\left(\frac{{\int }_0^{x^2}{\sec }^{2}tdt}{x\sin x}\right)$ is

• A. $3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

The correct option is C $1$

Given, $\underset{x\to 0}{lim}\left(\frac{{\int }_0^{x^2}{\sec }^{2}tdt}{x\sin x}\right)$
$=\underset{x\to 0}{lim}\frac{\frac{d}{dx}{\int }_0^{x^2}{\sec }^{2}tdt}{\frac{d}{dx}\left(x\sin x\right)}$
$=\underset{x\to 0}{lim}\frac{{\sec }^2{x}^2\cdot 2x}{\sin x+x\cos x}$ (by Leibnitz rule)
$=\underset{x\to 0}{lim}\frac{2x\cdot {\sec }^2{x}^2}{x(\frac{\sin x}{x}+\cos x)}$
$=\frac{2\times 1}{1+1}$ $(\because \underset{x\to 0}{lim}\frac{\sin x}{x}=1)$
$=1$