## Problem Statement

Calculate the limit of the function:

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

$\lim _{x \rightarrow 2}\left(\frac{\cos x}{\cos 2}\right)^{\frac{1}{x-2}}=\lim _{x \rightarrow 2}\left(e^{\ln \left(\frac{\cos x}{\cos 2}\right)}\right)^{\frac{1}{x-2}}=$

$=\lim _{x \rightarrow 2} e^{\frac{1}{x-2} \cdot \ln \left(\frac{\cos x}{\cos 2}\right)}=\exp \left\{\lim _{x \rightarrow 2} \frac{1}{x-2} \cdot \ln \left(\frac{\cos x}{\cos 2}\right)\right\}=$

Substitution:

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

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

We get:

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

Using the substitution of equivalent infinitesimals:

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

We get:

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

Using the substitution of equivalent infinitesimals:

![](https://cdn.mathpix.com/cropped/2024_05_22_41bbdfb6ea93bf7a2d4fg-13.jpg?height=141&width=697&top_left_y=2336&top_left_x=150)

We get:

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

$$
\begin{aligned}
& =\exp \left\{\lim _{y \rightarrow 0}-\sin \frac{y}{2}-\cos \frac{y}{2} \cdot \tan 2\right\}= \\
& =\exp \left\{-\sin \frac{0}{2}-\cos \frac{0}{2} \cdot \tan 2\right\}=\exp \{-0-1 \cdot \tan 2\}= \\
& =\exp \{-\tan 2\}=e^{-\tan 2}
\end{aligned}
$$

## Problem Kuznetsov Limits 19-4