## Condition of the problem

Calculate the limit of the function:

$$
\lim _{x \rightarrow 3}\left(\frac{9-2 x}{3}\right)^{\tan\left(\frac{\pi x}{6}\right)}
$$
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 3}\left(\frac{9-2 x}{3}\right)^{\operatorname{tg}\left(\frac{\pi x}{6}\right)}=\lim _{x \rightarrow 3}\left(3-\frac{2 x}{3}\right)^{\operatorname{tg}\left(\frac{\pi x}{6}\right)}= \\
& =\lim _{x \rightarrow 3}\left(e^{\ln \left(3-\frac{2 x}{3}\right)}\right)^{\operatorname{tg}\left(\frac{\pi x}{6}\right)}=\lim _{x \rightarrow 3}\left(e^{\operatorname{tg}\left(\frac{\pi x}{6}\right) \cdot \ln \left(3-\frac{2 x}{3}\right)}\right)= \\
& =\exp \left\{\lim _{x \rightarrow 3} \operatorname{tg}\left(\frac{\pi x}{6}\right) \cdot \ln \left(3-\frac{2 x}{3}\right)\right\}=
\end{aligned}
$$

Substitution:

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

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

We get:

$=\exp \left\{\lim _{y \rightarrow 0} \operatorname{tg}\left(\frac{\pi(y+3)}{6}\right) \cdot \ln \left(3-\frac{2(y+3)}{3}\right)\right\}=$

$=\exp \left\{\lim _{y \rightarrow 0} \operatorname{tg}\left(\frac{\pi y}{6}+\frac{\pi}{2}\right) \cdot \ln \left(3-\frac{2 y+6}{3}\right)\right\}=$

$=\exp \left\{\lim _{y \rightarrow 0}-\operatorname{ctg}\left(\frac{\pi y}{6}\right) \cdot \ln \left(1-\frac{2 y}{3}\right)\right\}=$

$=\exp \left\{\lim _{y \rightarrow 0}-\frac{1}{\operatorname{tg}\left(\frac{\pi y}{6}\right)} \cdot \ln \left(1-\frac{2 y}{3}\right)\right\}=$

Using the substitution of equivalent infinitesimals:

$\operatorname{tg}\left(\frac{\pi y}{6}\right) \sim \frac{\pi y}{6}$, as $y \rightarrow 0\left(\frac{\pi y}{6} \rightarrow 0\right)$

$\ln \left(1-\frac{2 y}{3}\right) \sim-\frac{2 y}{3}$, as $y \rightarrow 0\left(-\frac{2 y}{3} \rightarrow 0\right)$

We get:

$$
\begin{aligned}
& =\exp \left\{\lim _{y \rightarrow 0}-\frac{1}{\left(\frac{\pi y}{6}\right)} \cdot\left(-\frac{2 y}{3}\right)\right\}=\exp \left\{\lim _{y \rightarrow 0} \frac{4}{\pi}\right\}= \\
& =\exp \left\{\frac{4}{\pi}\right\}=e^{\frac{4}{\pi}}
\end{aligned}
$$

## Problem Kuznetsov Limits 19-15