## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow \pi} \frac{\sin 5 x}{\tan 3 x}$
## Solution

Substitution:

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

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

We get:

$\lim _{x \rightarrow \pi} \frac{\sin 5 x}{\tan 3 x}=\lim _{y \rightarrow 0} \frac{\sin 5(y+\pi)}{\tan 3(y+\pi)}=$

$=\lim _{y \rightarrow 0} \frac{\sin (5 y+5 \pi)}{\tan(3 y+3 \pi)}=\lim _{y \rightarrow 0} \frac{\sin (5 y+\pi)}{\tan(3 y+\pi)}=$

$=\lim _{y \rightarrow 0} \frac{-\sin 5 y}{\tan 3 y}=$

Using the substitution of equivalent infinitesimals:

$\sin 5 y \sim 5 y_{\text {, as }} y \rightarrow 0(5 y \rightarrow 0)$

$\tan 3 y \sim 3 y$, as $y \rightarrow 0(3 y \rightarrow 0)$

We get:

$=\lim _{y \rightarrow 0} \frac{-5 y}{3 y}=\lim _{y \rightarrow 0} \frac{-5}{3}=-\frac{5}{3}$

Problem Kuznetsov Limits 13-30