## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow \pi} \frac{\ln (\cos 2 x)}{\ln (\cos 4 x)}$
## Solution

Substitution:

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

We get:

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

Using the substitution of equivalent infinitesimals:

![](https://cdn.mathpix.com/cropped/2024_05_22_0cd048bfcc83fa1ff638g-09.jpg?height=109&width=1212&top_left_y=2122&top_left_x=164)

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

We get:

$=\lim _{y \rightarrow 0} \frac{-2 \sin ^{2} y}{-2 \sin ^{2} 2 y}=\lim _{y \rightarrow 0} \frac{\sin ^{2} y}{\sin ^{2} 2 y}=$

Using the substitution of equivalent infinitesimals:

![](https://cdn.mathpix.com/cropped/2024_05_22_0cd048bfcc83fa1ff638g-10.jpg?height=78&width=461&top_left_y=178&top_left_x=152)

$\sin 2 y \sim 2 y_{\text {, as }} y \rightarrow 0(2 y \rightarrow 0)$

We get:

$=\lim _{y \rightarrow 0} \frac{y^{2}}{(2 y)^{2}}=\lim _{y \rightarrow 0} \frac{y^{2}}{4 y^{2}}=\lim _{y \rightarrow 0} \frac{1}{4}=\frac{1}{4}$

## Problem Kuznetsov Limits 14-25