## Problem Statement

Calculate the limit of the function:

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

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

Substitution:

$x=y+1 \Rightarrow y=x-1$

$x \rightarrow 1 \Rightarrow y \rightarrow 0$

We get:

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

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

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

Using the substitution of equivalent infinitesimals:

$\ln \left(1-\frac{2 y}{y+1}\right) \sim-\frac{2 y}{y+1} \text{ as } y \rightarrow 0\left(-\frac{2 y}{y+1} \rightarrow 0\right)$

$\ln (1-y) \sim-y \text{ as } y \rightarrow 0(-y \rightarrow 0)$

We get:

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

$=\exp \left\{\frac{2}{0+1}\right\}=\exp \{2\}=e^{2}$

## Problem Kuznetsov Limits 19-23