## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 1}\left(\frac{e^{2 x}-e^{2}}{x-1}\right)^{x+1}$
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 1}\left(\frac{e^{2 x}-e^{2}}{x-1}\right)^{x+1}=\left(\lim _{x \rightarrow 1} \frac{e^{2 x}-e^{2}}{x-1}\right)^{\lim _{x \rightarrow 1} x+1}= \\
& =\left(\lim _{x \rightarrow 1} \frac{e^{2 x}-e^{2}}{x-1}\right)^{1+1}=\left(\lim _{x \rightarrow 1} \frac{e^{2}\left(e^{2 x-2}-1\right)}{x-1}\right)^{2}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$e^{2 x-2}-1 \sim 2 x-2$, as $x \rightarrow 1(2 x-2 \rightarrow 0)$

We get:

$$
=\left(\lim _{x \rightarrow 1} \frac{e^{2}(2 x-2)}{x-1}\right)^{2}=\left(\lim _{x \rightarrow 1} 2 e^{2}\right)^{2}=\left(2 e^{2}\right)^{2}=4 e^{4}
$$

## Problem Kuznetsov Limits 20-31