Answer
$$\lim_{x\to0}\frac{\sin^{-1}x}{x}=1$$
Work Step by Step
$$A=\lim_{x\to0}\frac{\sin^{-1}x}{x}$$
Because $\lim_{x\to0}(\sin^{-1}x)=\sin^{-1}0=0$ and $\lim_{x\to0}x=0,$
an indeterminate form of $\frac{0}{0}$ is here, so L'Hospital's Rule can be applied.
$$A=\lim_{x\to0}\frac{(\sin^{-1}x)'}{x'}$$
We have $(\sin^{-1}x)'=\frac{1}{\sqrt{1-x^2}}$, hence
$$A=\lim_{x\to0}\frac{\frac{1}{\sqrt{1-x^2}}}{1}$$
$$A=\lim_{x\to0}\frac{1}{\sqrt{1-x^2}}$$
$$A=\frac{1}{\sqrt{1-0^2}}$$
$$A=\frac{1}{\sqrt1}=1$$
