Answer
$$
\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{-2 x}}{\ln (x+1)}=4
$$
Work Step by Step
$$
\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{-2 x}}{\ln (x+1)}
$$
notice that as $ \quad x \rightarrow 0 \quad $ we have $ \quad ( e^{2 x}-e^{-2 x} ) \rightarrow 0 \quad $ and $ \quad \ln (x+1) \rightarrow 0 $
so, the limit is an indeterminate form of type $\frac{0}{0}$ so we can apply l’Hospital’s Rule:
$$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{e^{2 x}-e^{-2 x}}{\ln (x+1)} &
\quad\quad \rightarrow \frac{0}{0} \\
&\stackrel{\mathrm{H}}{=} \lim _{x \rightarrow 0} \frac{2 e^{2 x}+2 e^{-2 x}}{1 /(x+1)}\\
&=\frac{2+2}{1}\\
&=4
\end{aligned}
$$
