## Problem Statement

Calculate the limit of the function:

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

Substitution:

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

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

We get:

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

Using the substitution of equivalent infinitesimals:

$$
\tan \pi y \sim \pi y, \text { as } \pi y \rightarrow 0
$$

We get:

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

$$
=\lim _{y \rightarrow 0} \frac{y+1}{\pi\left(\sqrt{y^{2}+y+1}+1\right)}=\frac{0+1}{\pi\left(\sqrt{0^{2}+0+1}+1\right)}=\frac{1}{2 \pi}
$$

## Problem Kuznetsov Limits 13-8