## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow \pi} \frac{\left(x^{3}-\pi^{3}\right) \sin 5 x}{e^{\sin ^{2} x}-1}$
## Solution

Substitution:

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

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

We get:

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

Using the substitution of equivalent infinitesimals:

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

$\sin 5 y \sim 5 y$, as $y \rightarrow 0(5 y \rightarrow 0)$

We get:

$=\lim _{y \rightarrow 0} \frac{-y\left(y^{2}+3 y \pi+3 \pi^{2}\right) \cdot 5 y}{\sin ^{2} y}=$

Using the substitution of equivalent infinitesimals:

$\sin y \sim y$, as $y \rightarrow 0$

We get:

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{-y\left(y^{2}+3 y \pi+3 \pi^{2}\right) \cdot 5 y}{y^{2}}= \\
& =\lim _{y \rightarrow 0}-\left(y^{2}+3 y \pi+3 \pi^{2}\right) \cdot 5= \\
& =-\left(0^{2}+3 \cdot 0 \cdot \pi+3 \pi^{2}\right) \cdot 5=-15 \pi^{2}
\end{aligned}
$$

## Problem Kuznetsov Limits 14-23