Practicing Success
$\lim\limits_{x \rightarrow 0} \frac{1}{x} \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$ is |
2 1 0 none of these |
2 |
$\lim\limits_{x \rightarrow 0} \frac{1}{x} \sin ^{-1} \frac{2 x}{1+x^2}$, 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. |