## Problem Statement

Calculate the limit of the function:

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

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

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

Substitution:

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

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

We get:

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

Using the substitution of equivalent infinitesimals:

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

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

We get:

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

## Problem Kuznetsov Limits 19-27