Question

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

• A. $-2$
• B. $0$
• C. $2$
• D. $\infty$

Answer 1

Answer: (C)

$2$

Answer 2

Put $x=\tan \theta \Rightarrow \theta ={{\tan }^{-1}}x$
$\therefore$ $\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{x}{{\sin }^{-1}}\left( \frac{2\,\tan \theta }{1+{{\tan }^{2}}\theta } \right)$
$=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{x}\,{{\sin }^{-1}}(\sin 2\theta )$
$=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{x}\,.2\theta =\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\tan }^{-1}}x}{x}$
$=2\times \underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{\tan }^{-1}}x}{x}$
$=2\times 1=2$