## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow \frac{\pi}{2}}(\sin x)^{\frac{18 \sin x}{\operatorname{ctg} x}}$
## Solution

Substitution:

$$
\begin{aligned}
& x=y+\frac{\pi}{2} \Rightarrow y=x-\frac{\pi}{2} \\
& x \rightarrow \frac{\pi}{2} \Rightarrow y \rightarrow 0
\end{aligned}
$$

We obtain:

$$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{2}}(\sin x)^{\frac{18 \sin x}{\operatorname{ctg} x}}=\lim _{y \rightarrow 0}\left(\sin \left(y+\frac{\pi}{2}\right)\right)^{\frac{18 \sin \left(y+\frac{\pi}{2}\right)}{\operatorname{ctg}\left(y+\frac{\pi}{2}\right)}}= \\
& =\lim _{y \rightarrow 0}(\cos y)^{\frac{18 \cos y}{-\operatorname{tg} y}}=\lim _{y \rightarrow 0}\left(e^{\ln (\cos y)}\right)^{-\frac{18 \cos y}{\operatorname{tg} y}}= \\
& =\lim _{y \rightarrow 0} e^{-\frac{18 \cos y}{\operatorname{tg} y} \cdot \ln (\cos y)}=\exp \left\{\lim _{y \rightarrow 0}-\frac{18 \cos y}{\operatorname{tg} y} \cdot \ln (\cos y)\right\}= \\
& =\exp \left\{\lim _{y \rightarrow 0}-\frac{18 \cos y}{\operatorname{tg} y} \cdot \ln \left(1-2 \sin ^{2} \frac{y}{2}\right)\right\}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:
$\ln \left(1-2 \sin ^{2} \frac{y}{2}\right) \sim-2 \sin ^{2} \frac{y}{2}$, as $\quad y \rightarrow 0\left(-2 \sin ^{2} \frac{y}{2} \rightarrow 0\right)$ $\operatorname{tg} y \sim y_{\text {, as }} y \rightarrow 0$

We get:

$=\exp \left\{\lim _{y \rightarrow 0}-\frac{18 \cos y}{y} \cdot\left(-2 \sin ^{2} \frac{y}{2}\right)\right\}=$

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:

$$
\begin{aligned}
& =\exp \left\{\lim _{y \rightarrow 0} \frac{36 \cos y}{y} \cdot\left(\frac{y}{2}\right)^{2}\right\}=\exp \left\{\lim _{y \rightarrow 0} 9 \cdot y \cdot \cos y\right\}= \\
& =\exp \{9 \cdot 0 \cdot \cos 0\}=\exp \{0\}=e^{0}=1
\end{aligned}
$$

## Problem Kuznetsov Limits 19-28