Condition of the problem

Calculate the limit of the function:

$\lim _{x \rightarrow \pi} \frac{\operatorname{tg}\left(3^{\pi / x}-3\right)}{3^{\cos (3 x / 2)}-1}$
## Solution

Substitution:

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

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

We get:

$\lim _{x \rightarrow \pi} \frac{\operatorname{tg}\left(3^{\pi / x}-3\right)}{3^{\cos (3 x / 2)}-1}=\lim _{y \rightarrow 0} \frac{\operatorname{tg}\left(3^{\pi /(y+\pi)}-3\right)}{3^{\cos (3(y+\pi) / 2)}-1}=$

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{\operatorname{tg}\left(3^{1-y /(y+\pi)}-3\right)}{3^{\cos (3 y / 2+3 \pi / 2)}-1}=\lim _{y \rightarrow 0} \frac{\operatorname{tg}\left(3\left(3^{-y /(y+\pi)}-1\right)\right)}{3^{\sin (3 y / 2)}-1}= \\
& =\lim _{y \rightarrow 0} \frac{\operatorname{tg}\left(3\left(\left(e^{\ln 3}\right)^{-y /(y+\pi)}-1\right)\right)}{\left(e^{\ln 3}\right)^{\sin (3 y / 2)}-1}= \\
& =\lim _{y \rightarrow 0} \frac{\operatorname{tg}\left(3\left(e^{-y \cdot \ln 3 /(y+\pi)}-1\right)\right)}{e^{\ln 3 \cdot \sin (3 y / 2)}-1}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$e^{-y \cdot \ln 3 /(y+\pi)}-1 \sim-y \cdot \ln 3 /(y+\pi)$, as $y \rightarrow 0(-y \cdot \ln 3 /(y+\pi) \rightarrow 0)$

$e^{\ln 3 \cdot \sin (3 y / 2)}-1 \sim \ln 3 \cdot \sin (3 y / 2)$, as $y \rightarrow 0(\ln 3 \cdot \sin (3 y / 2) \rightarrow 0)$

We get:

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

Using the substitution of equivalent infinitesimals:

$\operatorname{tg}\left(-\frac{3 y \cdot \ln 3}{y+\pi}\right) \sim-\frac{3 y \cdot \ln 3}{y+\pi}$, as $\quad y \rightarrow 0\left(-\frac{3 y \cdot \ln 3}{y+\pi} \rightarrow 0\right)$

$\sin \frac{3 y}{2} \sim \frac{3 y}{2}_{\text {, as }} y \rightarrow 0\left(\frac{3 y}{2} \rightarrow 0\right)$

We get:

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{-\frac{3 y \cdot \ln 3}{y+\pi}}{\ln 3 \cdot \frac{3 y}{2}}=\lim _{y \rightarrow 0} \frac{-2 y}{(y+\pi) \cdot y}= \\
& =\lim _{y \rightarrow 0} \frac{-2}{y+\pi}=\frac{-2}{0+\pi}=-\frac{2}{\pi}
\end{aligned}
$$

## Problem Kuznetsov Limits 14-30