## Problem Statement

Calculate the limit of the function:

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

Substitution:

$$
\begin{aligned}
& x=y+\pi \Rightarrow y=x-\pi \\
& x \rightarrow \pi \Rightarrow y \rightarrow 0
\end{aligned}
$$

We get:

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

Using the substitution of equivalent infinitesimals:

$$
\sin \left(y \cdot \frac{y+2 \pi}{\pi}\right) \sim y \cdot \frac{y+2 \pi}{\pi}, \text { as } y \rightarrow 0\left(y \cdot \frac{y+2 \pi}{\pi} \rightarrow 0\right)
$$

$$
\begin{aligned}
& e^{\ln 2(\sqrt{1-\sin y}-1)}-1 \sim \ln 2(\sqrt{1-\sin y}-1) \\
& y \rightarrow 0(\ln 2(\sqrt{1-\sin y}-1) \rightarrow 0)
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{-y \cdot \frac{y+2 \pi}{\pi}}{2 \ln 2(\sqrt{1-\sin y}-1)}= \\
& =\lim _{y \rightarrow 0} \frac{-y \cdot \frac{y+2 \pi}{\pi}(\sqrt{1-\sin y}+1)}{2 \ln 2(\sqrt{1-\sin y}-1)(\sqrt{1-\sin y}+1)}= \\
& =\lim _{y \rightarrow 0} \frac{-y \cdot \frac{y+2 \pi}{\pi}(\sqrt{1-\sin y}+1)}{2 \ln 2(1-\sin y-1)}= \\
& =\lim _{y \rightarrow 0} \frac{y \cdot \frac{y+2 \pi}{\pi}(\sqrt{1-\sin y}+1)}{2 \sin y \cdot \ln 2}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

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

We get:

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{y \cdot \frac{y+2 \pi}{\pi}(\sqrt{1-\sin y}+1)}{2 y \cdot \ln 2}=\lim _{y \rightarrow 0} \frac{\frac{y+2 \pi}{\pi}(\sqrt{1-\sin y}+1)}{2 \ln 2}= \\
& =\frac{\frac{0+2 \pi}{\pi}(\sqrt{1-\sin 0}+1)}{2 \ln 2}=\frac{2(\sqrt{1}+1)}{2 \ln 2}=\frac{2}{\ln 2}
\end{aligned}
$$

## Problem Kuznetsov Limits 14-31