## Problem Statement

Calculate the limit of the function:

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

## Substitution:

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

We obtain:

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

Using the substitution of equivalent infinitesimals:

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

We get:

$=\lim _{y \rightarrow 0} \frac{1-\cos y}{(-\sin y \cdot \ln 3)^{2}}=\lim _{y \rightarrow 0} \frac{1-\cos y}{\sin ^{2} y \cdot \ln ^{2} 3}=$

Using the substitution of equivalent infinitesimals:
$\sin y \sim y$, as $y \rightarrow 0$

$1-\cos y \sim \frac{y^{2}}{2}$, as $y \rightarrow 0$

We get:

$$
=\lim _{y \rightarrow 0} \frac{\frac{y^{2}}{2}}{y^{2} \cdot \ln ^{2} 3}=\lim _{y \rightarrow 0} \frac{1}{2 \ln ^{2} 3}=\frac{1}{2 \ln ^{2} 3}
$$

## Problem Kuznetsov Limits 14-22