## Problem Statement

Calculate the limit of the numerical sequence:

$\lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{2}}-\sqrt{n^{2}+5}}{\sqrt[5]{n^{7}}-\sqrt{n+1}}$
## Solution

$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{2}}-\sqrt{n^{2}+5}}{\sqrt[5]{n^{7}}-\sqrt{n+1}}=\lim _{n \rightarrow \infty} \frac{n^{-\frac{7}{5}}\left(\sqrt[3]{n^{2}}-\sqrt{n^{2}+5}\right)}{n^{-\frac{7}{5}}\left(\sqrt[5]{n^{7}}-\sqrt{n+1}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{2} \cdot n^{-\frac{21}{5}}}-\sqrt{n^{2} \cdot n^{-\frac{14}{5}}+5 n^{-\frac{14}{5}}}}{\sqrt[5]{n^{7} \cdot n^{-7}}-\sqrt{n \cdot n^{-\frac{14}{5}}}+n^{-\frac{14}{5}}}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{-\frac{11}{5}}}-\sqrt{n^{-\frac{4}{5}}+5 n^{-\frac{14}{5}}}}{\sqrt[5]{1}-\sqrt{n^{-\frac{9}{5}}+n^{-\frac{14}{5}}}}= \\
& =\frac{\sqrt[3]{0}-\sqrt{0+5 \cdot 0}}{1-\sqrt{0+0}}=\frac{0}{1}=0
\end{aligned}
$$

Problem Kuznetsov Limits 4-22