Question

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

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

Answer 1

Answer: (C)

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