## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow \frac{\pi}{4}}(\tan x)^{1 / \cos \left(\frac{3 \pi}{4}-x\right)}$
## Solution

$\lim _{x \rightarrow \frac{\pi}{4}}(\operatorname{tg} x)^{1 / \cos \left(\frac{3 \pi}{4}-x\right)}=\lim _{x \rightarrow \frac{\pi}{4}}\left(e^{\ln (\operatorname{tg} x)}\right)^{1 / \cos \left(\frac{3 \pi}{4}-x\right)}=$

$=\lim _{x \rightarrow \frac{\pi}{4}} e^{\ln (\operatorname{tg} x) / \cos \left(\frac{3 \pi}{4}-x\right)}=\exp \left\{\lim _{x \rightarrow \frac{\pi}{4}} \ln (\operatorname{tg} x) / \cos \left(\frac{3 \pi}{4}-x\right)\right\}=$

Substitution:

$x=y+\frac{\pi}{4} \Rightarrow y=x-\frac{\pi}{4}$

$x \rightarrow \frac{\pi}{4} \Rightarrow y \rightarrow 0$

We get:

$$
\begin{aligned}
& =\exp \left\{\lim _{y \rightarrow 0} \ln \left(\operatorname{tg}\left(y+\frac{\pi}{4}\right)\right) / \cos \left(\frac{3 \pi}{4}-\left(y+\frac{\pi}{4}\right)\right)\right\}= \\
& =\exp \left\{\lim _{y \rightarrow 0} \ln \left(\frac{\operatorname{tg} y+\operatorname{tg} \frac{\pi}{4}}{1-\operatorname{tg} y \cdot \operatorname{tg} \frac{\pi}{4}}\right) / \cos \left(\frac{3 \pi}{4}-y-\frac{\pi}{4}\right)\right\}= \\
& =\exp \left\{\lim _{y \rightarrow 0} \ln \left(\frac{\operatorname{tg} y+1}{1-\operatorname{tg} y}\right) / \cos \left(\frac{\pi}{2}-y\right)\right\}= \\
& =\exp \left\{\lim _{y \rightarrow 0} \ln \left(1+\frac{2 \operatorname{tg} y}{1-\operatorname{tg} y}\right) /-\sin (-y)\right\}= \\
& =\exp \left\{\lim _{y \rightarrow 0} \frac{1}{\sin y} \cdot \ln \left(1+\frac{2 \operatorname{tg} y}{1-\operatorname{tg} y}\right)\right\}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

![](https://cdn.mathpix.com/cropped/2024_05_22_3e5254fbaf58ae3260d1g-12.jpg?height=155&width=1255&top_left_y=2606&top_left_x=150)
$\sin y \sim y_{\text {, as }} y \rightarrow 0$

We get:

$=\exp \left\{\lim _{y \rightarrow 0} \frac{1}{y} \cdot \frac{2 \operatorname{tg} y}{1-\operatorname{tg} y}\right\}=$

Using the substitution of equivalent infinitesimals:

$\operatorname{tg} y \sim y_{\text {, as }} y \rightarrow 0$

We get:

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

## Problem Kuznetsov Limits 19-6