## Problem Statement

Calculate the limit of the function:

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

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

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

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

We get:

$$
\begin{aligned}
& =\left(\lim _{y \rightarrow 0} \frac{e^{\sin \pi(y+1)}-1}{(y+1)-1}\right)^{2}=\left(\lim _{y \rightarrow 0} \frac{e^{\sin (\pi y+\pi)}-1}{y}\right)^{2}= \\
& =\left(\lim _{y \rightarrow 0} \frac{e^{-\sin \pi y}-1}{y}\right)^{2}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$e^{-\sin \pi y}-1 \sim -\sin \pi y$, as $y \rightarrow 0(-\sin \pi y \rightarrow 0)$

We get:

$$
=\left(\lim _{y \rightarrow 0} \frac{-\sin \pi y}{y}\right)^{2}=
$$

Using the substitution of equivalent infinitesimals:

$\sin \pi y \sim \pi y$, as $y \rightarrow 0(\pi y \rightarrow 0)$

We get:

$=\left(\lim _{y \rightarrow 0} \frac{-\pi y}{y}\right)^{2}=(-\pi)^{2}=\pi^{2}$

## Problem Kuznetsov Limits 20-24