Answer
$$
\lim _{x \rightarrow 0} \frac{e^{x}-1}{\tan x}=1
$$
Work Step by Step
$$
\lim _{x \rightarrow 0} \frac{e^{x}-1}{\tan x}
$$
notice that as $ \quad x \rightarrow 0 \quad $ we have $ \quad e^{x}-1 \rightarrow 0 \quad $ and $ \quad \tan x \rightarrow 0 $
so that we have an indeterminate form of type $\frac{0}{0}$ and by l’Hospital’s Rule we gives:
$$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{e^{x}-1}{\tan x} & \quad\quad \rightarrow \frac{0}{0}\\
& \stackrel{\mathrm{H}}{=} \lim _{x \rightarrow 0} \frac{e^{x}}{\sec ^{2} x}\\
&=\frac{1}{1}\\
&=1
\end{aligned}
$$
