## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 1}\left(\frac{2 x-1}{x}\right)^{\frac{\ln (3+2 x)}{\ln (2-x)}}$
## Solution

$\lim _{x \rightarrow 1}\left(\frac{2 x-1}{x}\right)^{\frac{\ln (3+2 x)}{\ln (2-x)}}=\lim _{x \rightarrow 1}\left(e^{\ln \left(\frac{2 x-1}{x}\right)}\right)^{\frac{\ln (3+2 x)}{\ln (2-x)}}=$

$=\lim _{x \rightarrow 1} e^{\frac{\ln (3+2 x)}{\ln (2-x)} \cdot \ln \left(\frac{2 x-1}{x}\right)}=\exp \left\{\lim _{x \rightarrow 1} \frac{\ln (3+2 x)}{\ln (2-x)} \cdot \ln \left(\frac{2 x-1}{x}\right)\right\}=$

Substitution:

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

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

We get:

$=\exp \left\{\lim _{y \rightarrow 0} \frac{\ln (3+2(y+1))}{\ln (2-(y+1))} \cdot \ln \left(\frac{2(y+1)-1}{y+1}\right)\right\}=$

$=\exp \left\{\lim _{y \rightarrow 0} \frac{\ln (5+2 y)}{\ln (1-y)} \cdot \ln \left(\frac{2 y+1}{y+1}\right)\right\}=$

$=\exp \left\{\lim _{y \rightarrow 0} \frac{\ln (5+2 y)}{\ln (1-y)} \cdot \ln \left(1+\frac{y}{y+1}\right)\right\}=$

Using the substitution of equivalent infinitesimals:

$\ln \left(1+\frac{y}{y+1}\right) \sim \frac{y}{y+1}, \quad y \rightarrow 0\left(\frac{y}{y+1} \rightarrow 0\right)$

$\ln (1-y) \sim -y$, as $y \rightarrow 0(-y \rightarrow 0)$

We get:

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

## Problem Kuznetsov Limits 19-31