## Problem Statement

Calculate the limit of the numerical sequence:

$$
\lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{3}+5}-\sqrt{3 n^{4}+2}}{1+3+5+\ldots+(2 n-1)}
$$
## Solution

$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{3}+5}-\sqrt{3 n^{4}+2}}{1+3+5+\ldots+(2 n-1)}=\lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{3}+5}-\sqrt{3 n^{4}+2}}{\frac{(1+(2 n-1)) n}{2}}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{3}+5}-\sqrt{3 n^{4}+2}}{\frac{2 n \cdot n}{2}}=\lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{3}+5}-\sqrt{3 n^{4}+2}}{n^{2}}= \\
& =\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(\sqrt[3]{n^{3}+5}-\sqrt{3 n^{4}+2}\right)}{\frac{1}{n^{2}} \cdot n^{2}}=\lim _{n \rightarrow \infty} \frac{\sqrt[3]{\frac{1}{n^{3}}+\frac{5}{n^{6}}}-\sqrt{3+\frac{2}{n^{4}}}}{1}= \\
& =\frac{\sqrt[3]{0+0}-\sqrt{3+0}}{1}=-\sqrt{3}
\end{aligned}
$$

## Problem Kuznetsov Limits 6-15