Answer
$$
\lim _{x \rightarrow 0} \frac{\sin ^{-1} x}{x}=1
$$
Work Step by Step
$$
\lim _{x \rightarrow 0} \frac{\sin ^{-1} x}{x}
$$
Since
$$
\lim _{x \rightarrow 0}( \sin ^{-1} x)=0
$$
and
$$
\lim _{x \rightarrow 0}( x)=0
$$
the limit is an indeterminate form of type $\frac{0}{0}$ so we can apply l’Hospital’s Rule:
$$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\sin ^{-1} x}{x} & \stackrel{\text { H }}{=} \lim _{x \rightarrow 0} \frac{1 / \sqrt{1-x^{2}}}{1} \\
&=\lim _{x \rightarrow 0} \frac{1}{\sqrt{1-x^{2}}} \\
&=\frac{1}{1} \\
&=1
\end{aligned}
$$
$$$$
