## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow \pi} \frac{e^{\pi}-e^{x}}{\sin 5 x-\sin 3 x}$
## Solution

$\lim _{x \rightarrow \pi} \frac{e^{\pi}-e^{x}}{\sin 5 x-\sin 3 x}=\lim _{x \rightarrow \pi} \frac{-e^{\pi}\left(e^{x-\pi}-1\right)}{2 \sin \frac{5 x-3 x}{2} \cos \frac{5 x+3 x}{2}}=$

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

Substitution:

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

$x \rightarrow \pi \Rightarrow y \rightarrow 0$

We get:

$=\lim _{y \rightarrow 0} \frac{-e^{\pi}\left(e^{y+\pi-\pi}-1\right)}{2 \sin (y+\pi) \cos 4(y+\pi)}=\lim _{y \rightarrow 0} \frac{-e^{\pi}\left(e^{y}-1\right)}{-2 \sin y(-\cos (4 y+4 \pi))}=$

$=\lim _{y \rightarrow 0} \frac{e^{\pi}\left(e^{y}-1\right)}{2 \sin y \cos 4 y}=$

Using the substitution of equivalent infinitesimals:

$\sin y \sim y$, as $y \rightarrow 0$

$e^{y}-1 \sim y$, as $y \rightarrow 0$

We get:

$=\lim _{y \rightarrow 0} \frac{e^{\pi} y}{2 y \cos 4 y}=\lim _{y \rightarrow 0} \frac{e^{\pi}}{2 \cos 4 y}=\frac{e^{\pi}}{2 \cos (4 \cdot 0)}=\frac{e^{\pi}}{2}$

## Problem Kuznetsov Limits 13-19