## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 0}\left(\frac{1+\tan x \cdot \cos 2 x}{1+\tan x \cdot \cos 5 x}\right)^{\frac{1}{x^{3}}}
$$
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(\frac{1+\tan x \cdot \cos 2 x}{1+\tan x \cdot \cos 5 x}\right)^{\frac{1}{x^{3}}}= \\
& =\lim _{x \rightarrow 0}\left(e^{\ln \left(\frac{1+\tan x \cdot \cos 2 x}{1+\tan x \cdot \cos 5 x}\right)}\right)^{\frac{1}{x^{3}}}= \\
& =\lim _{x \rightarrow 0} e^{\frac{1}{x^{3}} \cdot \ln \left(\frac{1+\tan x \cdot \cos 2 x}{1+\tan x \cdot \cos 5 x}\right)}= \\
& =\exp \left\{\lim _{x \rightarrow 0} \frac{1}{x^{3}} \cdot \ln \left(\frac{1+\tan x \cdot \cos 2 x}{1+\tan x \cdot \cos 5 x}\right)\right\}= \\
& =\exp \left\{\lim _{x \rightarrow 0} \frac{1}{x^{3}} \cdot \ln \left(\frac{1+\tan x \cdot \cos 5 x-\tan x \cdot \cos 5 x+\tan x \cdot \cos 2 x}{1+\tan x \cdot \cos 5 x}\right)\right\}= \\
& =\exp \left\{\lim _{x \rightarrow 0} \frac{1}{x^{3}} \cdot \ln \left(1-\frac{\tan x \cdot \cos 5 x-\tan x \cdot \cos 2 x}{1+\tan x \cdot \cos 5 x}\right)\right\}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$\ln \left(1-\frac{\tan x \cdot \cos 5 x-\tan x \cdot \cos 2 x}{1+\tan x \cdot \cos 5 x}\right) \sim-\frac{\tan x \cdot \cos 5 x-\tan x \cdot \cos 2 x}{1+\tan x \cdot \cos 5 x}$, as $x \rightarrow 0\left(-\frac{\tan x \cdot \cos 5 x-\tan x \cdot \cos 2 x}{1+\tan x \cdot \cos 5 x} \rightarrow 0\right)$

We get:

$$
\begin{aligned}
& =\exp \left\{\lim _{x \rightarrow 0}-\frac{1}{x^{3}} \cdot \frac{\tan x \cdot \cos 5 x-\tan x \cdot \cos 2 x}{1+\tan x \cdot \cos 5 x}\right\}= \\
& =\exp \left\{\lim _{x \rightarrow 0}-\frac{\tan x}{x^{3}} \cdot \frac{\cos 5 x-\cos 2 x}{1+\tan x \cdot \cos 5 x}\right\}= \\
& =\exp \left\{\lim _{x \rightarrow 0}-\frac{\tan x}{x^{3}} \cdot \frac{-2 \sin \frac{5 x+2 x}{2} \sin \frac{5 x-2 x}{2}}{1+\tan x \cdot \cos 5 x}\right\}=
\end{aligned}
$$

$=\exp \left\{\lim _{x \rightarrow 0}-\frac{\tan x}{x^{3}} \cdot \frac{-2 \sin \frac{7 x}{2} \sin \frac{3 x}{2}}{1+\tan x \cdot \cos 5 x}\right\}=$

Using the substitution of equivalent infinitesimals:

$\tan x \sim x$, as $x \rightarrow 0$

![](https://cdn.mathpix.com/cropped/2024_05_22_fcfb644d65eb046b31c7g-12.jpg?height=146&width=784&top_left_y=515&top_left_x=150)

$\sin \frac{3 x}{2} \sim \frac{3 x}{2}$, as $x \rightarrow 0\left(\frac{3 x}{2} \rightarrow 0\right)$

We get:

$=\exp \left\{\lim _{x \rightarrow 0}-\frac{x}{x^{3}} \cdot \frac{-2 \cdot \frac{7 x}{2} \cdot \frac{3 x}{2}}{1+\tan x \cdot \cos 5 x}\right\}=$

$=\exp \left\{\lim _{x \rightarrow 0} \frac{2 \cdot \frac{7}{2} \cdot \frac{3}{2}}{1+\tan x \cdot \cos 5 x}\right\}=$

$=\exp \left\{\lim _{x \rightarrow 0} \frac{21}{2(1+\tan x \cdot \cos 5 x)}\right\}=$

$=\exp \left\{\frac{21}{2(1+\tan 0 \cdot \cos (5 \cdot 0))}\right\}=$

$=\exp \left\{\frac{21}{2}\right\}=e^{\frac{21}{2}}$

## Problem Kuznetsov Limits 17-26