A) 3
B) 2
C) 1
D) -1
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\int_{0}^{{{x}^{2}}}{{{\sec }^{2}}t\,dr}}{x\sin x}\] \[\left( \frac{0}{0}form \right)\] Applying L' Hospital rule, \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sec }^{2}}{{x}^{2}}.\,\,2x}{x\cos x+\sin x}\] Since, \[\frac{d}{dx}\int_{0}^{{{x}^{2}}}{{{\sec }^{2}}t\,.\,dt={{\sec }^{2}}{{x}^{2}}.\,\,2x}\] (by Leibnitz rule) Again, applying L' Hospital rule, \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2x\,.\,2{{\sec }^{2}}{{x}^{2}}.\,\,\tan {{x}^{2}}.\,2x+2se{{c}^{2}}{{x}^{2}}}{-x\sin x+\cos x+\cos x}\] \[=\frac{0+2{{\sec }^{2}}0}{0+2\cos 0}=1\]
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