## Problem Statement

Calculate the limit of the numerical sequence:

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

$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\sqrt{n(n+2)}-\sqrt{n^{2}-2 n+3}\right)= \\
& =\lim _{n \rightarrow \infty} \frac{\left(\sqrt{n(n+2)}-\sqrt{n^{2}-2 n+3}\right)\left(\sqrt{n(n+2)}+\sqrt{n^{2}-2 n+3}\right)}{\sqrt{n(n+2)}+\sqrt{n^{2}-2 n+3}}= \\
& =\lim _{n \rightarrow \infty} \frac{n(n+2)-\left(n^{2}-2 n+3\right)}{\sqrt{n(n+2)}+\sqrt{n^{2}-2 n+3}}=\lim _{n \rightarrow \infty} \frac{n^{2}+2 n-n^{2}+2 n-3}{\sqrt{n(n+2)}+\sqrt{n^{2}-2 n+3}}= \\
& =\lim _{n \rightarrow \infty} \frac{4 n-3}{\sqrt{n(n+2)}+\sqrt{n^{2}-2 n+3}}= \\
& =\lim _{n \rightarrow \infty} \frac{\frac{1}{n}(4 n-3)}{\frac{1}{n\left(\sqrt{n(n+2)}+\sqrt{n^{2}-2 n+3}\right)}}= \\
& =\lim _{n \rightarrow \infty} \frac{4-\frac{3}{n}}{\sqrt{1+\frac{2}{n}}+\sqrt{1-\frac{2}{n}+\frac{3}{n^{2}}}}=\frac{4-0}{\sqrt{1+0}+\sqrt{1-0+0}}=\frac{4}{2}=2
\end{aligned}
$$

## Problem Kuznetsov Limits 5-8