## Problem Statement

Calculate the limit of the function:

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

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

Substitution:

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

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

We get:

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

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

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

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:

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

## Problem Kuznetsov Limits 19-29