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. $1/2$

Answer 1

Answer: (C)

$2$

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

can be written as

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

it is of the form $\frac{0}{0}$

So, we will apply L Hospital rule until we will get determinate form

Differentiating numerator and denominator we will get,

$={lim}_{x\to 0}\frac{\frac{1}{\sqrt{1-(\frac{2x}{1+x^2}{)}^2}}.\frac{2(1+x^2)-4x^2}{(1+x^2{)}^2}}{1}={lim}_{x\to 0}\frac{1}{\sqrt{1-(\frac{2x}{1+x^2}{)}^2}}.\frac{2(1+x^2)-4x^2}{(1+x^2{)}^2}=2$