## Problem Statement

Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{4}}\left(\frac{\ln (\operatorname{tg} x)}{1-\operatorname{ctg} x}\right)^{1 /\left(x+\frac{\pi}{4}\right)}$
## Solution

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

$=\left(\lim _{x \rightarrow \frac{\pi}{4}} \frac{\ln (\operatorname{tg} x)}{1-\operatorname{ctg} x}\right)^{1 /\left(\frac{\pi}{4}+\frac{\pi}{4}\right)}=\left(\lim _{x \rightarrow \frac{\pi}{4}} \frac{\ln (\operatorname{tg} x)}{1-\operatorname{ctg} x}\right)^{1 /\left(\frac{\pi}{2}\right)}=$

$=\left(\lim _{x \rightarrow \frac{\pi}{4}} \frac{\ln (\operatorname{tg} x)}{1-\operatorname{ctg} x}\right)^{\frac{2}{\pi}}=\left(\lim _{x \rightarrow \frac{\pi}{4}} \frac{\ln (1-(1-\operatorname{tg} x))}{1-\operatorname{ctg} x}\right)^{\frac{2}{\pi}}=$

Using the substitution of equivalent infinitesimals:

$\ln (1-(1-\operatorname{tg} x)) \sim-(1-\operatorname{tg} x)_{\text {, as }} x \rightarrow \frac{\pi}{4}(-(1-\operatorname{tg} x) \rightarrow 0)$

We get:

$$
\begin{aligned}
& =\left(\lim _{x \rightarrow \frac{\pi}{4}} \frac{-(1-\operatorname{tg} x)}{1-\operatorname{ctg} x}\right)^{\frac{2}{\pi}}=\left(\lim _{x \rightarrow \frac{\pi}{4}} \frac{\operatorname{tg} x-1}{1-\operatorname{ctg} x}\right)^{\frac{2}{\pi}}= \\
& =\left(\lim _{x \rightarrow \frac{\pi}{4}} \frac{\frac{1}{\operatorname{ctg} x}-\frac{\operatorname{ctg} x}{\operatorname{ctg} x}}{1-\operatorname{ctg} x}\right)^{\frac{2}{\pi}}=\left(\lim _{x \rightarrow \frac{\pi}{4}} \frac{\frac{1}{\operatorname{ctg} x}(1-\operatorname{ctg} x)}{1-\operatorname{ctg} x}\right)^{\frac{2}{\pi}}= \\
& =\left(\lim _{x \rightarrow \frac{\pi}{4}} \operatorname{tg} x\right)^{\frac{2}{\pi}}=\left(\operatorname{tg} \frac{\pi}{4}\right)^{\frac{2}{\pi}}=1^{\frac{2}{\pi}}=1
\end{aligned}
$$

Limits 20-3