The value of $\lim _{x \rightarrow 0} \frac{\int_0^{x^2} \sec ^2 t d t}{x \sin x}$ is

The value of $\lim _{x \rightarrow 0} \frac{\int_0^{x^2} \sec ^2 t d t}{x \sin x}$ is
  1. 0
  2. 3
  3. 2
  4. 1

Solution

$\operatorname{Lim}_{x \rightarrow 0} \frac{\frac{d}{d x} \int_0^{x^2} \sec ^2 t d t}{\frac{d}{d x}(x \sin x)}=\operatorname{Lim}_{x \rightarrow 0} \frac{\sec ^2 x^2 \cdot 2 x}{\sin x+x \cos x}$ (by L'Hospital rule) $\operatorname{Lim}_{x \rightarrow 0} \frac{2 \sec ^2 x^2}{\left(\frac{\sin x}{x}+\cos x\right)}=\frac{2 \times 1}{1+1}=1$

Asked in: JEE Main 2003

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