Condition of the problem

Calculate the limit of the function:

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

Substitution:

$$
\begin{aligned}
& x=y+1 \Rightarrow y=x-1 \\
& x \rightarrow 1 \Rightarrow y \rightarrow 0
\end{aligned}
$$

We get:

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

Using the substitution of equivalent infinitesimals:

$\ln (1+y) \sim y_{, \text {as }} y \rightarrow 0$

We get:

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

## Problem Kuznetsov Limits 13-2