Question

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

Answer 1

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

$=\underset{x\to 0}{lim}$ $\frac{x+\frac{1^2}{3!}{x}^3+\frac{1^2{3}^2}{5!}{x}^5..............}{x}$

$=\underset{x\to 0}{lim}$ $\frac{x(1+\frac{1^2}{3!}{x}^2+\frac{1^2{3}^2}{5!}{x}^4..............)}{x}$

$=\underset{x\to 0}{lim}$ $1+\frac{1^2}{3!}{x}^2+\frac{1^2{3}^2}{5!}{x}^4+........$

$=$ $1+\frac{1^2}{3!}{0}^2+\frac{1^2{3}^2}{5!}{0}^4+........$

$=$ $1$