## Problem Statement

Calculate the limit of the numerical sequence:

$\lim _{n \rightarrow \infty}\left(\frac{n+4}{n+2}\right)^{n}$
## Solution

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

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

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

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

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

$=e^{\lim _{n \rightarrow \infty} \frac{2}{1+\frac{2}{n}}}=e^{\frac{2}{1+0}}=e^{2}$

## Problem Kuznetsov Limits $7-24$