## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 2} \frac{\operatorname{arctg}\left(x^{2}-2 x\right)}{\sin 3 \pi x}$
## Solution

Substitution:

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

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

We get:

$\lim _{x \rightarrow 2} \frac{\operatorname{arctg}\left(x^{2}-2 x\right)}{\sin 3 \pi x}=\lim _{y \rightarrow 0} \frac{\operatorname{arctg}\left((y+2)^{2}-2(y+2)\right)}{\sin 3 \pi(y+2)}=$

$=\lim _{y \rightarrow 0} \frac{\operatorname{arctg}\left(y^{2}+4 y+4-2 y-4\right)}{\sin (3 \pi y+6 \pi)}=$

$=\lim _{y \rightarrow 0} \frac{\operatorname{arctg}\left(y^{2}+2 y\right)}{\sin 3 \pi y}=$

Using the substitution of equivalent infinitesimals:

$\operatorname{arctg}\left(y^{2}+2 y\right) \sim\left(y^{2}+2 y\right)_{\text {, as }} y \rightarrow 0\left(\left(y^{2}+2 y\right) \rightarrow 0\right)$

$\sin 3 \pi y \sim 3 \pi y{ }_{\text {, as }} y \rightarrow 0(3 \pi y \rightarrow 0)$

We get:

$=\lim _{y \rightarrow 0} \frac{y^{2}+2 y}{3 \pi y}=\lim _{y \rightarrow 0} \frac{y+2}{3 \pi}=\frac{0+2}{3 \pi}=\frac{2}{3 \pi}$

## Problem Kuznetsov Limits 13-26