## Problem Statement

Calculate the limit of the numerical sequence:

$\lim _{n \rightarrow \infty} \frac{n \sqrt[4]{11 n}+\sqrt{25 n^{4}-81}}{(n-7 \sqrt{n}) \sqrt{n^{2}-n+1}}$
## Solution

$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{n \sqrt[4]{11 n}+\sqrt{25 n^{4}-81}}{(n-7 \sqrt{n}) \sqrt{n^{2}-n+1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(n \sqrt[4]{11 n}+\sqrt{25 n^{4}-81}\right)}{\frac{1}{n^{2}}(n-7 \sqrt{n}) \sqrt{n^{2}-n+1}}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt[4]{\frac{11}{n^{3}}}+\sqrt{25-\frac{81}{n^{4}}}}{\left(\frac{1}{n}(n-7 \sqrt{n})\right)\left(\frac{1}{n} \sqrt{n^{2}-n+1}\right)}=\lim _{n \rightarrow \infty} \frac{\sqrt[4]{\frac{11}{n^{3}}}+\sqrt{25-\frac{81}{n^{4}}}}{\left(1-7 \sqrt{\frac{1}{n}}\right) \sqrt{1-\frac{1}{n}+\frac{1}{n^{2}}}}= \\
& =\frac{\sqrt[4]{0}+\sqrt{25-0}}{(1-7 \sqrt{0}) \sqrt{1-0+0}}=\frac{5}{1}=5
\end{aligned}
$$

## Problem Kuznetsov Limits 4-21