## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 3} \frac{\log _{3} x-1}{\tan \pi x}$
## Solution

$\lim _{x \rightarrow 3} \frac{\log _{3} x-1}{\tan \pi x}=\lim _{x \rightarrow 3} \frac{\log _{3} x-\log _{3} 3}{\tan \pi x}=$
$=\lim _{x \rightarrow 3} \frac{\log _{3} \frac{x}{3}}{\tan \pi x}=\lim _{x \rightarrow 3} \frac{\ln \frac{x}{3} / \ln 3}{\tan \pi x}=$

$=\lim _{x \rightarrow 3} \frac{\ln \frac{x}{3}}{\ln 3 \cdot \tan \pi x}=$

Substitution:

$x=y+3 \Rightarrow y=x-3$

$x \rightarrow 3 \Rightarrow y \rightarrow 0$

We get:

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{\ln \frac{y+3}{3}}{\ln 3 \cdot \tan \pi(y+3)}=\lim _{y \rightarrow 0} \frac{\ln \left(1+\frac{y}{3}\right)}{\ln 3 \cdot \tan(\pi y+3 \pi)}= \\
& =\lim _{y \rightarrow 0} \frac{\ln \left(1+\frac{y}{3}\right)}{\ln 3 \cdot \tan(\pi y+\pi)}=\lim _{y \rightarrow 0} \frac{\ln \left(1+\frac{y}{3}\right)}{\ln 3 \cdot \tan \pi y}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$\ln \left(1+\frac{y}{3}\right) \sim \frac{y}{3}$ as $y \rightarrow 0\left(\frac{y}{3} \rightarrow 0\right)$

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

We get:

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

## Problem Kuznetsov Limits 16-30