## Problem Statement

Calculate the limit of the function:

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

Substitution:

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

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

We get:

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

Using the substitution of equivalent infinitesimals:

$1-\cos \pi y \sim \frac{(\pi y)^{2}}{2}$, as $y \rightarrow 0$

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

We get:

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

## Problem Kuznetsov Limits 13-5