## Problem Statement

Calculate the limit of the function:
$\lim _{x \rightarrow \pi} \frac{\cos 5 x-\cos 3 x}{\sin ^{2} x}$
## Solution

$\lim _{x \rightarrow \pi} \frac{\cos 5 x-\cos 3 x}{\sin ^{2} x}=\lim _{x \rightarrow \pi} \frac{-2 \sin \frac{5 x+3 x}{2} \sin \frac{5 x-3 x}{2}}{\sin ^{2} x}=$

$=\lim _{x \rightarrow \pi} \frac{-2 \sin 4 x \sin x}{\sin ^{2} x}=\lim _{x \rightarrow \pi} \frac{-2 \sin 4 x}{\sin x}=$

Substitution:

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

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

We get:

$=\lim _{y \rightarrow 0} \frac{-2 \sin 4(y+\pi)}{\sin (y+\pi)}=\lim _{y \rightarrow 0} \frac{-2 \sin (4 y+4 \pi)}{-\sin y}=$

$=\lim _{y \rightarrow 0} \frac{2 \sin 4 y}{\sin y}=$

Using the substitution of equivalent infinitesimals:

$\sin 4 y \sim 4 y_{\text {, as }} y \rightarrow 0(4 y \rightarrow 0)$

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

$=\lim _{y \rightarrow 0} \frac{2 \cdot 4 y}{y}=\lim _{y \rightarrow 0} \frac{8}{1}=8$

## Problem Kuznetsov Limits 13-9