## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow \pi} \frac{\cos 3 x-\cos x}{\tan^{2} 2 x}$
## Solution

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

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

Substitution:

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

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

We get:

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

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

Using the substitution of equivalent infinitesimals:

$\sin y \sim y$, as $y \rightarrow 0$
$\sin 2 y \sim 2 y$, as $y \rightarrow 0$ (and $2 y \rightarrow 0$)

$\tan 2 y \sim 2 y$, as $y \rightarrow 0$ (and $2 y \rightarrow 0$)

We get:

$$
=\lim _{y \rightarrow 0} \frac{2 \cdot 2 y \cdot y}{(2 y)^{2}}=\lim _{y \rightarrow 0} \frac{4 y^{2}}{4 y^{2}}=\lim _{y \rightarrow 0} 1=1
$$

## Problem Kuznetsov Limits 13-31