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