## Problem Statement

Calculate the limit of the numerical sequence:

$\lim _{n \rightarrow \infty}\left(\frac{n+3}{n+5}\right)^{n+4}$
## Solution

$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{n+3}{n+5}\right)^{n+4}=\lim _{n \rightarrow \infty}\left(\frac{n+5}{n+3}\right)^{-n-4}= \\
& =\lim _{n \rightarrow \infty}\left(\frac{n+3+2}{n+3}\right)^{-n-4}=\lim _{n \rightarrow \infty}\left(1+\frac{2}{n+3}\right)^{-n-4}= \\
& =\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{n+3}{2}\right)}\right)^{-n-4}= \\
& =\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{n+3}{2}\right)}\right)^{\left(\frac{n+3}{2}\right)\left(\frac{2}{n+3}\right)(-n-4)}= \\
& =\left(\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{n+3}{2}\right)}\right)^{\left(\frac{n+3}{2}\right)}\right)^{\lim _{n \rightarrow \infty}\left(\frac{2}{n+3}\right)(-n-4)}= \\
& =\{\text { Using the second remarkable limit }\}= \\
& =e^{\lim _{n \rightarrow \infty} \frac{2(-n-4)}{n+3}}=e^{\lim _{n \rightarrow \infty} \frac{\frac{1}{n} 2(-n-4)}{\frac{1}{n}(n+3)}}= \\
& =e^{\lim _{n \rightarrow \infty} \frac{2\left(-1-\frac{4}{n}\right)}{1+\frac{3}{n}}}=e^{\frac{2(-1-0)}{1+0}}=e^{-2}
\end{aligned}
$$

Problem Kuznetsov Limits 7-17