## Problem Statement

Calculate the limit of the function:

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

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

$x=y+1 \Rightarrow y=x-1$

$x \rightarrow 1 \Rightarrow y \rightarrow 0$

We get:

$=(1+e)^{\lim _{y \rightarrow 0} \frac{\sin \pi(y+1)}{1-(y+1)}}=\left(1+e e^{\lim _{y \rightarrow 0} \frac{\sin (\pi y+\pi)}{-y}}=\right.$
$=(1+e)^{\lim _{y \rightarrow 0} \frac{-\sin \pi y}{-y}}=(1+e)^{\lim _{y \rightarrow 0} \frac{\sin \pi y}{y}}=$

Using the substitution of equivalent infinitesimals:

$\sin \pi y \sim \pi y \text{ as } y \rightarrow 0 (\pi y \rightarrow 0)$

We get:

$$
=(1+e)^{\lim _{y \rightarrow 0} \frac{\pi y}{y}}=(1+e)^{\lim _{y \rightarrow 0} \pi}=(1+e)^{\pi}
$$

## Kuznetsov Limits 20-9