## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan 3x}{\tan x}$
## Solution

Substitution:

$x=y+\frac{\pi}{2} \Rightarrow y=x-\frac{\pi}{2}$

$x \rightarrow \frac{\pi}{2} \Rightarrow y \rightarrow 0$

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan 3 x}{\tan x}=\lim _{y \rightarrow 0} \frac{\tan 3\left(y+\frac{\pi}{2}\right)}{\tan \left(y+\frac{\pi}{2}\right)}=\lim _{y \rightarrow 0} \frac{\tan \left(3 y+\frac{3 \pi}{2}\right)}{\tan \left(y+\frac{\pi}{2}\right)}= \\
& =\lim _{y \rightarrow 0} \frac{-\cot(3 y)}{-\cot y}=\lim _{y \rightarrow 0} \frac{\tan y}{\tan 3 y}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

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

We get:

$=\lim _{y \rightarrow 0} \frac{y}{3 y}=\lim _{y \rightarrow 0} \frac{1}{3}=\frac{1}{3}$

## Problem Kuznetsov Limits 13-6