## Problem Statement

Calculate the limit of the numerical sequence:

$\lim _{n \rightarrow \infty} \frac{n \sqrt{71 n}-\sqrt[3]{64 n^{6}+9}}{(n-\sqrt[3]{n}) \sqrt{11+n^{2}}}$
## Solution

$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n \sqrt{71 n}-\sqrt[3]{64 n^{6}+9}}{(n-\sqrt[3]{n}) \sqrt{11+n^{2}}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(n \sqrt{71 n}-\sqrt[3]{64 n^{6}+9}\right)}{\frac{1}{n^{2}}(n-\sqrt[3]{n}) \sqrt{11+n^{2}}}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt{\frac{71}{n}}-\sqrt[3]{64+\frac{9}{n^{6}}}}{\left(\frac{1}{n}(n-\sqrt[3]{n})\right)\left(\frac{1}{n} \sqrt{11+n^{2}}\right)}=\lim _{n \rightarrow \infty} \frac{\sqrt{\frac{71}{n}}-\sqrt[3]{64+\frac{9}{n^{6}}}}{\left(1-\sqrt[3]{\frac{1}{n^{2}}}\right) \sqrt{\frac{11}{n^{2}}+1}}= \\
& =\frac{\sqrt{0}-\sqrt[3]{64+0}}{(1-\sqrt[3]{0}) \sqrt{0+1}}=\frac{-4}{1}=-4
\end{aligned}
$$

## Problem Kuznetsov Limits 4-26