## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow \pi}\left(\operatorname{ctg}\left(\frac{x}{4}\right)\right)^{1 / \cos \left(\frac{x}{2}\right)}$
## Solution

Substitution:

$$
\begin{aligned}
& x=4 y+\pi \Rightarrow y=\frac{1}{4}(x-\pi) \\
& x \rightarrow \pi \Rightarrow y \rightarrow 0
\end{aligned}
$$

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow \pi}\left(\operatorname{ctg}\left(\frac{4 y+\pi}{4}\right)\right)^{1 / \cos \left(\frac{4 y+\pi}{2}\right)}=\lim _{y \rightarrow 0}\left(\operatorname{ctg}\left(y+\frac{\pi}{4}\right)\right)^{1 / \cos \left(2 y+\frac{\pi}{2}\right)}= \\
& =\lim _{y \rightarrow 0}\left(\operatorname{ctg}\left(y+\frac{\pi}{4}\right)\right)^{-\frac{1}{\sin 2 y}}=\lim _{y \rightarrow 0}\left(e^{\ln \left(\operatorname{ctg}\left(y+\frac{\pi}{4}\right)\right)}\right)^{-\frac{1}{\sin 2 y}}= \\
& =\lim _{y \rightarrow 0} e^{-\frac{1}{\sin 2 y} \cdot \ln \left(\operatorname{ctg}\left(y+\frac{\pi}{4}\right)\right)}= \\
& =\exp \left\{\lim _{y \rightarrow 0}-\frac{1}{\sin 2 y} \cdot \ln \left(\operatorname{ctg}\left(y+\frac{\pi}{4}\right)\right)\right\}= \\
& =\exp \left\{\lim _{y \rightarrow 0}-\frac{1}{\sin 2 y} \cdot \ln \left(\frac{1}{\operatorname{tg}\left(y+\frac{\pi}{4}\right)}\right)\right\}= \\
& =\exp \left\{\lim _{y \rightarrow 0}-\frac{1}{\sin 2 y} \cdot \ln \left(\frac{1}{\left(\frac{\operatorname{tg} y+\operatorname{tg} \frac{\pi}{4}}{1-\operatorname{tg} y \cdot \operatorname{tg} \frac{\pi}{4}}\right)}\right)\right\}=
\end{aligned}
$$

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

Using the substitution of equivalent infinitesimals:

$\ln \left(1-\frac{2 \operatorname{tg} y}{\operatorname{tg} y+1}\right) \sim-\frac{2 \operatorname{tg} y}{1-\operatorname{tg} y}$, as $\quad y \rightarrow 0\left(-\frac{2 \operatorname{tg} y}{1-\operatorname{tg} y} \rightarrow 0\right)$ $\sin 2 y \sim 2 y_{\text {, as }} y \rightarrow 0(2 y \rightarrow 0)$

We get:

$=\exp \left\{\lim _{y \rightarrow 0}-\frac{1}{2 y} \cdot \frac{-2 \operatorname{tg} y}{\operatorname{tg} y+1}\right\}=$

Using the substitution of equivalent infinitesimals:

$\operatorname{tg} y \sim y_{\text {, as }} y \rightarrow 0$

We get:

$$
\begin{aligned}
& =\exp \left\{\lim _{y \rightarrow 0} \frac{1}{y} \cdot \frac{y}{\operatorname{tg} y+1}\right\}=\exp \left\{\lim _{y \rightarrow 0} \frac{1}{\operatorname{tg} y+1}\right\}= \\
& =\exp \left\{\frac{1}{\operatorname{tg} 0+1}\right\}=\exp \left\{\frac{1}{0+1}\right\}=\exp \{1\}=e
\end{aligned}
$$

## Problem Kuznetsov Limits 19-30