Question

Solve:-
$\underset{x\to 0}{lim}\frac{{\sin }^{-1}x}{x}$

Answer 1

$\underset{x\to 0}{lim}\frac{{\sin }^{-1}x}{x}$

$=\frac{{\sin }^{-1}0}{0}=\frac{0}{0}$ form.

Using L'Hospitals Rule,

$=\underset{x\to 0}{lim}\frac{\frac{1}{\sqrt{1-x^2}}}{1}$

$=\underset{x\to 0}{lim}\frac{1}{\sqrt{1-x^2}}=\frac{1}{\sqrt{1-0}}=\frac{1}{1}=1$