## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 2 \pi} \frac{(x-2 \pi)^{2}}{\tan(\cos x-1)}
$$
## Solution

Substitution:

$x=y+2 \pi \Rightarrow y=x-2 \pi$

$x \rightarrow 2 \pi \Rightarrow y \rightarrow 0$

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow 2 \pi} \frac{(x-2 \pi)^{2}}{\operatorname{tg}(\cos x-1)}=\lim _{y \rightarrow 0} \frac{((y+2 \pi)-2 \pi)^{2}}{\operatorname{tg}(\cos (y+2 \pi)-1)}= \\
& =\lim _{y \rightarrow 0} \frac{y^{2}}{\operatorname{tg}(\cos y-1)}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$\operatorname{tg}(\cos y-1) \sim(\cos y-1)_{, \text {as }} y \rightarrow 0((\cos y-1) \rightarrow 0)$

We get:

$=\lim _{y \rightarrow 0} \frac{y^{2}}{\cos y-1}=$

Using the substitution of equivalent infinitesimals:

$1-\cos y \sim{\frac{y^{2}}{2}}_{, \text {as }} y \rightarrow 0$

We get:

$=\lim _{y \rightarrow 0} \frac{y^{2}}{-\frac{y^{2}}{2}}=\lim _{y \rightarrow 0} \frac{1}{-\frac{1}{2}}=-2$