## Problem Statement

Calculate the limit of the numerical sequence:

$$
\lim _{n \rightarrow \infty}\left(\frac{n^{3}+n+1}{n^{3}+2}\right)^{2 n^{2}}
$$
## Solution

$\lim _{n \rightarrow \infty}\left(\frac{n^{3}+n+1}{n^{3}+2}\right)^{2 n^{2}}=\lim _{n \rightarrow \infty}\left(\frac{n^{3}+2+n-1}{n^{3}+2}\right)^{2 n^{2}}=$

$$
\begin{aligned}
& =\lim _{n \rightarrow \infty}\left(1+\frac{n-1}{n^{3}+2}\right)^{2 n^{2}}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{n^{3}+2}{n-1}\right)}\right)^{2 n^{2}}= \\
& =\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{n^{3}+2}{n-1}\right)}\right)^{\left(\frac{n^{3}+2}{n-1}\right)\left(\frac{n-1}{n^{3}+2}\right)\left(2 n^{2}\right)}= \\
& =\left(\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{n^{3}+2}{n-1}\right)}\right)^{\left(\frac{n^{3}+2}{n-1}\right)}\right)^{\lim _{n \rightarrow \infty}\left(\frac{n-1}{n^{3}+2}\right)\left(2 n^{2}\right)}= \\
& =\{\text { Using the second remarkable limit }\}= \\
& =e^{\lim _{n \rightarrow \infty}\left(\frac{n-1}{n^{3}+2}\right)\left(2 n^{2}\right)}=e^{\lim ^{n \rightarrow \infty} \frac{2 n^{3}-2 n^{2}}{n^{3}+2}}= \\
& =e^{\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{3}}\left(2 n^{3}-2 n^{2}\right)}{\frac{1}{n^{3}}\left(n^{3}+2\right)}}=e^{\lim _{n \rightarrow \infty} \frac{2-\frac{2}{n}}{1+\frac{2}{n^{3}}}}= \\
& =e^{\frac{2-0}{1+0}}=e^{2}
\end{aligned}
$$

## Problem Kuznetsov Limits $7-27$