## Problem Statement

Calculate the limit of the function:

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

$\lim _{x \rightarrow 3}\left(\frac{\sin x}{\sin 3}\right)^{\frac{1}{x-3}}=\lim _{x \rightarrow 3}\left(e^{\ln \left(\frac{\sin x}{\sin 3}\right)}\right)^{\frac{1}{x-3}}=$

$=\lim _{x \rightarrow 3} e^{\frac{1}{x-3} \cdot \ln \left(\frac{\sin x}{\sin 3}\right)}=\exp \left\{\lim _{x \rightarrow 3} \frac{1}{x-3} \cdot \ln \left(\frac{\sin x}{\sin 3}\right)\right\}=$

Substitution:

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

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

We get:

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

Using the substitution of equivalent infinitesimals:

$$
\begin{aligned}
& \ln \left(1+2 \sin \frac{y}{2}\left(\cos \frac{y}{2} \cdot \cot 3-\sin \frac{y}{2}\right)\right) \sim 2 \sin \frac{y}{2}\left(\cos \frac{y}{2} \cdot \cot 3-\sin \frac{y}{2}\right), \text { as } \\
& y \rightarrow 0\left(2 \sin \frac{y}{2}\left(\cos \frac{y}{2} \cdot \cot 3-\sin \frac{y}{2}\right) \rightarrow 0\right)
\end{aligned}
$$

We get:

$$
=\exp \left\{\lim _{y \rightarrow 0} \frac{1}{y} \cdot 2 \sin \frac{y}{2}\left(\cos \frac{y}{2} \cdot \cot 3-\sin \frac{y}{2}\right)\right\}=
$$

Using the substitution of equivalent infinitesimals:

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

We get:

$=\exp \left\{\lim _{y \rightarrow 0} \frac{1}{y} \cdot 2 \cdot \frac{y}{2} \cdot\left(\cos \frac{y}{2} \cdot \cot 3-\sin \frac{y}{2}\right)\right\}=$

$=\exp \left\{\lim _{y \rightarrow 0}\left(\cos \frac{y}{2} \cdot \cot 3-\sin \frac{y}{2}\right)\right\}=$

$=\exp \left\{\cos \frac{0}{2} \cdot \cot 3-\sin \frac{0}{2}\right\}=\exp \{1 \cdot \cot 3-0\}=$

$=\exp \{\cot 3\}=e^{\cot 3}$

## Problem Kuznetsov Limits 19-26