## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow \frac{\pi}{2}}(1+\cos 3 x)^{\sec x}$
## Solution

$\lim _{x \rightarrow \frac{\pi}{2}}(1+\cos 3 x)^{\sec x}=\lim _{x \rightarrow \frac{\pi}{2}}(1+\cos 3 x)^{\frac{1}{\cos x}}=$

Substitution:

$x=y+\frac{\pi}{2} \Rightarrow y=x-\frac{\pi}{2}$

$x \rightarrow \frac{\pi}{2} \Rightarrow y \rightarrow 0$

We get:

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

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

Using the substitution of equivalent infinitesimals:

$\ln (1+\sin 3 y) \sim \sin 3 y_{\text {, as }} y \rightarrow 0(\sin 3 y \rightarrow 0)$

$\sin y \sim y$, as $y \rightarrow 0$

We get:

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

Using the substitution of equivalent infinitesimals:

$\sin 3 y \sim 3 y_{\text {, as }} y \rightarrow 0(3 y \rightarrow 0)$

We get:

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

## Problem Kuznetsov Limits 19-20