$\lim\limits_{x \rightarrow 0} \frac{1}{x} \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$ is
Answer & explanation
Correct answer: option 1
$\lim\limits_{x \rightarrow 0} \frac{1}{x} \sin ^{-1} \frac{2 x}{1+x^2}$, let $x=\cos \theta$
$\Rightarrow \lim\limits_{x \rightarrow 0} \frac{1}{x} \sin ^{-1} (\sin 2 \theta)=\lim\limits_{x \rightarrow 0} \frac{1}{x} 2 \theta$
$=2 \lim\limits_{x \rightarrow 0} \frac{1}{x} \tan ^{-1} x=2 \lim\limits_{x \rightarrow 0} \frac{\tan ^{-1} x}{x}=2$
Hence (1) is the correct answer.