## Problem Statement

Calculate the limit of the function:

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

Substitution:

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

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

We get:

$\lim _{x \rightarrow 1} \frac{x^{2}-1}{\ln x}=\lim _{y \rightarrow 0} \frac{(y+1)^{2}-1}{\ln (y+1)}=$

Using the substitution of equivalent infinitesimals:

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

We get:

$=\lim _{y \rightarrow 0} \frac{y^{2}+2 y+1-1}{y}=\lim _{y \rightarrow 0} \frac{y^{2}+2 y}{y}=\lim _{y \rightarrow 0}(y+2)=0+2=2$

Problem Kuznetsov Limits 13-1