A) \[-2\]
B) \[0\]
C) \[2\]
D) \[\infty \]
Correct Answer: C
Solution :
Put \[x=\tan \theta \Rightarrow \theta ={{\tan }^{-1}}x\] \[\therefore \] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{x}{{\sin }^{-1}}\left( \frac{2\,\tan \theta }{1+{{\tan }^{2}}\theta } \right)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{x}\,{{\sin }^{-1}}(\sin 2\theta )\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{x}\,.2\theta =\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\tan }^{-1}}x}{x}\] \[=2\times \underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{\tan }^{-1}}x}{x}\] \[=2\times 1=2\]
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