Answer
$1$
Work Step by Step
We have
$$
\lim _{x \rightarrow 0}\frac{e^x-1}{\sin x}=\frac{0}{0}
$$
which is an intermediate form, so we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow 0}\frac{e^x-1}{\sin x}= \lim _{x \rightarrow 0}\frac{e^x}{\cos x}=1
$$
