## Problem Statement

Calculate the limit of the numerical sequence:

$\lim _{n \rightarrow \infty}\left(\frac{2 n^{2}+2}{2 n^{2}+1}\right)^{n^{2}}$
## Solution

$\lim _{n \rightarrow \infty}\left(\frac{2 n^{2}+2}{2 n^{2}+1}\right)^{n^{2}}=\lim _{n \rightarrow \infty}\left(\frac{2 n^{2}+1+1}{2 n^{2}+1}\right)^{n^{2}}=$

$=\lim _{n \rightarrow \infty}\left(1+\frac{1}{2 n^{2}+1}\right)^{n^{2}}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{2 n^{2}+1}\right)^{\left(2 n^{2}+1\right) \frac{1}{2 n^{2}+1} n^{2}}=$

$=\left(\lim _{n \rightarrow \infty}\left(1+\frac{1}{2 n^{2}+1}\right)^{\left(2 n^{2}+1\right)}\right)^{\lim _{n \rightarrow \infty} \frac{n^{2}}{2 n^{2}+1}}=$

$=\{$ Use the second remarkable limit $\}=$

$=e^{\lim _{n \rightarrow \infty} \frac{n^{2}}{2 n^{2}+1}}=e^{\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}} n^{2}}{\frac{1}{n^{2}}\left(2 n^{2}+1\right)}}=e^{\lim _{n \rightarrow \infty} \frac{1}{2+\frac{1}{n^{2}}}}=$

$=e^{\frac{1}{2+0}}=e^{\frac{1}{2}}=\sqrt{e}$

## Problem Kuznetsov Limits 7-5