A) 3
B) 2
C) 1
D) 0
Correct Answer: C
Solution :
Given, \[\underset{h\to 0}{\mathop{\lim }}\,\left( \frac{\int_{0}^{{{x}^{2}}}{{{\sec }^{2}}t\,dt}}{x\sin x} \right)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{d}{dx}\int_{0}^{{{x}^{2}}}{{{\sec }^{2}}t\,dt}}{\frac{d}{dx}(x\sin x)}\] (by L?Hospital's rule) \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sec }^{2}}{{x}^{2}}.2x}{\sin x+x\cos x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2x.{{\sec }^{2}}{{x}^{2}}}{\left( \frac{\sin x}{x}+\cos x \right)}\] \[\left[ \because \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}=1 \right]\] \[=\frac{2\times 1}{1+1}=1\]
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