Condition of the problem

$$
\lim _{n \rightarrow \infty} \frac{(1+2 n)^{3}-8 n^{5}}{(1+2 n)^{2}+4 n^{2}}
$$
## Solution

$$
\begin{aligned}
& \lim _{n \rightarrow x} \frac{(1+2 n)^{3}-8 n^{5}}{(1+2 n)^{2}+4 n^{2}}=\lim _{n \rightarrow x} \frac{1+6 n+12 n^{2}+8 n^{3}-8 n^{5}}{1+4 n+4 n^{2}+4 n^{2}}= \\
& =\lim _{n \rightarrow x} \frac{1+6 n+12 n^{2}}{1+4 n+8 n^{2}}=\lim _{n \rightarrow x} \frac{n^{2}\left(\frac{1}{n^{2}}+\frac{6}{n}+12\right)}{n^{2}\left(\frac{1}{n^{2}}+\frac{4}{n}+8\right)}= \\
& =\lim _{n \rightarrow x} \frac{\frac{1}{n^{2}}+\frac{6}{n}+12}{\frac{1}{n^{2}}+\frac{4}{n}+8}=\frac{0+0+12}{0+8}=\frac{12}{8}=1.5
\end{aligned}
$$

## Problem Kuznetsov Limits 3-7