## Problem Statement

Calculate the limit of the numerical sequence:

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

$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(n+1)^{4}-(n-1)^{4}}{(n+1)^{3}+(n-1)^{3}}=\lim _{n \rightarrow \infty} \frac{\left((n+1)^{2}-(n-1)^{2}\right) \cdot\left((n+1)^{2}+(n-1)^{2}\right)}{(n+1)^{3}+(n-1)^{3}}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(n^{2}+2 n+1-n^{2}+2 n-1\right) \cdot\left(n^{2}+2 n+1+n^{2}-2 n+1\right)}{(n+1)^{3}+(n-1)^{3}}= \\
& =\lim _{n \rightarrow \infty} \frac{4 n\left(2 n^{2}+2\right)}{(n+1)^{3}+(n-1)^{3}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{3}} 8 n\left(n^{2}+1\right)}{n^{3}\left((n+1)^{3}+(n-1)^{3}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{8\left(1+\frac{1}{n^{2}}\right)}{\left(1+\frac{1}{n}\right)^{3}+\left(1-\frac{1}{n}\right)^{3}}=\frac{8 \cdot 1}{1^{3}+1^{3}}=4
\end{aligned}
$$

## Problem Kuznetsov Limits 3-24