Question

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

• A. 1
• B. 0
• C. 2
• D. $\frac{1}{2}$

Answer 1

The correct option is C 2

Let $L=\underset{x\to 0}{lim}\frac{1}{x}si{n}^{-1}\left(\frac{2x}{1+x^2}\right)$
Now let $x=\tan \theta$ so as $x\to 0,\theta \to 0$
$\therefore L=\underset{\theta \to 0}{lim}\frac{1}{\tan \theta }si{n}^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)=\underset{\theta \to 0}{lim}\frac{1}{\tan \theta }si{n}^{-1}2\theta$
$\Rightarrow L=2\underset{\theta \to 0}{lim}\frac{\theta }{\tan \theta }=2$