## Problem Statement

Calculate the limit of the numerical sequence:

$$
\lim _{n \rightarrow \infty}\left(\sqrt{n^{2}-3 n+2}-n\right)
$$
## Solution

$\lim _{n \rightarrow \infty}\left(\sqrt{n^{2}-3 n+2}-n\right)=\lim _{n \rightarrow \infty} \frac{\left(\sqrt{n^{2}-3 n+2}-n\right)\left(\sqrt{n^{2}-3 n+2}+n\right)}{\sqrt{n^{2}-3 n+2}+n}=$
$=\lim _{n \rightarrow \infty} \frac{n^{2}-3 n+2-n^{2}}{\sqrt{n^{2}-3 n+2}+n}=\lim _{n \rightarrow \infty} \frac{-3 n+2}{\sqrt{n^{2}-3 n+2}+n}=$
$=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}(-3 n+2)}{\frac{1}{n}\left(\sqrt{n^{2}-3 n+2}+n\right)}=\lim _{n \rightarrow \infty} \frac{-3+\frac{2}{n}}{\sqrt{1-\frac{3}{n}+\frac{2}{n^{2}}}+1}=$
$=\frac{-3+0}{\sqrt{1-0+0}+1}=\frac{-3}{2}=-\frac{3}{2}$

## Problem Kuznetsov Limits 5-6