## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 2 \pi}(\cos x)^{\frac{\operatorname{ctg} 2 x}{\sin 3 x}}$
## Solution

$\lim _{x \rightarrow 2 \pi}(\cos x)^{\frac{\operatorname{ctg} 2 x}{\sin 3 x}}=\lim _{x \rightarrow 2 \pi}\left(e^{\ln (\cos x)}\right)^{\frac{\operatorname{ctg} 2 x}{\sin 3 x}}=$

$=\lim _{x \rightarrow 2 \pi} e^{\frac{\operatorname{ctg} 2 x}{\sin 3 x} \cdot \ln (\cos x)}=\exp \left\{\lim _{x \rightarrow 2 \pi} \frac{\operatorname{ctg} 2 x}{\sin 3 x} \cdot \ln (\cos x)\right\}=$

Substitution:

$x=y+2 \pi \Rightarrow y=x-2 \pi$

$x \rightarrow 2 \pi \Rightarrow y \rightarrow 0$

We get:

$=\exp \left\{\lim _{y \rightarrow 0} \frac{\operatorname{ctg} 2(y+2 \pi)}{\sin 3(y+2 \pi)} \cdot \ln (\cos (y+2 \pi))\right\}=$

$=\exp \left\{\lim _{y \rightarrow 0} \frac{\operatorname{ctg}(2 y+4 \pi)}{\sin (3 y+6 \pi)} \cdot \ln (\cos y)\right\}=$

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

Using the substitution of equivalent infinitesimals:

$\operatorname{tg} 2 y \sim 2 y$, as $y \rightarrow 0(2 y \rightarrow 0)$

$\sin 3 y \sim 3 y$, as $y \rightarrow 0(3 y \rightarrow 0)$

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

We get:

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

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:

$$
\begin{aligned}
& =\exp \left\{\lim _{y \rightarrow 0} \frac{1}{3 y^{2}} \cdot\left(-\left(\frac{y}{2}\right)^{2}\right)\right\}=\exp \left\{\lim _{y \rightarrow 0} \frac{1}{3 y^{2}} \cdot\left(-\frac{y^{2}}{4}\right)\right\}= \\
& =\exp \left\{\lim _{y \rightarrow 0}-\frac{1}{12}\right\}=e^{-\frac{1}{12}}
\end{aligned}
$$

## Problem Kuznetsov Limits 19-9