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Q. $ \underset{x\to 0}{\mathop{\lim }}\,\frac{1}{x}{{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right) $ =

J & K CETJ & K CET 2007Limits and Derivatives

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 $