Answer
$4e^4$
Work Step by Step
We have
$$
\lim _{x \rightarrow 2}\frac{e^{x^2}-e^4}{x-2}=\frac{0}{0}
$$
is an intermediate form, then we can apply L’Hôpital’s Rule as follows
$$\lim _{x \rightarrow 2}\frac{e^{x^2}-e^4}{x-2}=\lim _{x \rightarrow 2}\frac{2xe^{x^2}}{1}=4e^4.$$
