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
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0
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3
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2
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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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