Condition of the problem

Calculate the limit of the numerical sequence:

$$
\lim _{n \rightarrow \infty} \frac{\sqrt{n-1}-\sqrt{n^{2}+1}}{\sqrt[3]{3 n^{3}+3}+\sqrt[3]{n^{5}+1}}
$$
## Solution

$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt{n-1}-\sqrt{n^{2}+1}}{\sqrt[3]{3 n^{3}+3}+\sqrt{n^{5}+1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(\sqrt{n-1}-\sqrt{n^{2}+1}\right)}{\frac{1}{n}\left(\sqrt[3]{3 n^{3}+3}+\sqrt[1]{n^{5}+1}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt{\frac{1}{n}-\frac{1}{n^{2}}}-\sqrt{1+\frac{1}{n^{2}}}}{\sqrt[5]{3+\frac{3}{n^{3}}}+\sqrt[1]{n+\frac{1}{n^{3}}}}=\left\{\frac{\sqrt{0-0}-\sqrt{1+0}}{\sqrt[3]{3+0}+\sqrt[4]{\infty}}=\frac{-1}{\infty}\right\}=-\infty
\end{aligned}
$$

## Problem Kuznetsov Limits 4-2

Calculate the limit of the numerical sequence:

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

## Solution

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

## Problem Kuznetsov Limits 5-2