Question

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

• (A) 3
• (B) 2
• (C) 1
• (D) 0

Answer 1

The Correct Option is C

Explanation:
Given:The expression $\underset{x\to 0}{lim}\frac{{\int }_0^{x^2}{\sec }^{2}tdt}{x\sin x}$We have to evaluate the given expression.Here, we'll use fundamental integrals of trigonometric functions.$$\begin{gathered} \underset{x\to 0}{lim}\frac{{\int }_0^{x^2}{\sec }^{2}tdt}{x\sin x}[\int {\sec }^{2}tdt=\tan t] \\ \underset{x\to 0}{lim}\frac{[\tan t]x^2}{x\sin x} \end{gathered}$$$=\underset{x\to 0}{lim}\frac{\tan x^2}{x\sin x}$$=\underset{x\to 0}{lim}\frac{\frac{\tan x^2}{x^2}}{\frac{x^{2}x}{x}}[\text{Dividing by}{x}^2\text{in Numerator \& denominator}]$$\frac{\underset{x\to 0}{lim}\left(\frac{\tan x^2}{x^2}\right)}{\frac{\sin x}{x}}=\frac{1}{1}=1$[Using limit of trigonometric functions ]Hence, the correct option is (C).