## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 1} \frac{1-x^{2}}{\sin \pi x}$
## Solution

Substitution: u

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

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

We get:

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

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

Using the substitution of equivalent infinitesimals:

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

We get:

$=\lim _{y \rightarrow 0} \frac{y^{2}+2 y}{\pi y}=\lim _{y \rightarrow 0} \frac{y+2}{\pi}=\frac{0+2}{\pi}=\frac{2}{\pi}$

## Problem Kuznetsov Limits 13-27