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