## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 1} \frac{3^{5 x-3}-3^{2 x^{2}}}{\tan \pi x}
$$
## Solution

$\lim _{x \rightarrow 1} \frac{3^{5 x-3}-3^{2 x^{2}}}{\tan \pi x}=\lim _{x \rightarrow 1} \frac{3^{2 x^{2}}\left(3^{5 x-3-2 x^{2}}-1\right)}{\tan \pi x}=$[^0]$x=y+1 \Rightarrow y=x-1$

$x \rightarrow 1 \Rightarrow y \rightarrow 0$

We get:

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{3^{2(y+1)^{2}}\left(3^{5(y+1)-3-2(y+1)^{2}}-1\right)}{\tan \pi(y+1)}= \\
& =\lim _{y \rightarrow 0} \frac{3^{2(y+1)^{2}}\left(3^{5 y+5-3-2\left(y^{2}+2 y+1\right)}-1\right)}{\tan(\pi y+\pi)}= \\
& =\lim _{y \rightarrow 0} \frac{3^{2(y+1)^{2}}\left(3^{5 y+2-2 y^{2}-4 y-2}-1\right)}{\tan \pi y}= \\
& =\lim _{y \rightarrow 0} \frac{3^{2(y+1)^{2}}\left(3^{y-2 y^{2}}-1\right)}{\tan \pi y}=\lim _{y \rightarrow 0} \frac{3^{2(y+1)^{2}}\left(\left(e^{\ln 3}\right)^{y-2 y^{2}}-1\right)}{\tan \pi y}= \\
& =\lim _{y \rightarrow 0} \frac{3^{2(y+1)^{2}}\left(e^{\left(y-2 y^{2}\right) \ln 3}-1\right)}{\tan \pi y}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

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

$\tan \pi y \sim \pi y$, as $\pi y \rightarrow 0$

We get:

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{3^{2(y+1)^{2}}\left(y-2 y^{2}\right) \ln 3}{\pi y}=\lim _{y \rightarrow 0} \frac{3^{2(y+1)^{2}}(1-2 y) \ln 3}{\pi}= \\
& =\frac{3^{2(0+1)^{2}}(1-2 \cdot 0) \ln 3}{\pi}=\frac{3^{2} \cdot 1 \cdot \ln 3}{\pi}=\frac{9 \ln 3}{\pi}
\end{aligned}
$$

## Problem Kuznetsov Limits 13-15