Question

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

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

Answer 1

The correct option is C

$2$

Explanation for the correct option:

Expanding the given equation and applying the limits:

$$\begin{gathered} \underset{x\to 0}{\lim }\left(\frac{1}{x}\right){\sin }^{-1}\left(\frac{2x}{1+x^2}\right) \\ =\underset{x\to 0}{\lim }\left(\frac{1}{x}\right){\sin }^{-1}\left(\sin 2{\tan }^{-1}x\right)\left[\frac{d\left({\tan }^{-1}x\right)}{dx}=\frac{1}{1+x^2}\right] \\ =\underset{x\to 0}{\lim }\frac{2{\tan }^{-1}x}{x} \end{gathered}$$

Applying the limits,

$=2\left[\underset{x\to 0}{\lim }\left(\frac{{\tan }^{-1}x}{x}\right)=1\right]$

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

Therefore, the correct answer is option (C).