## Problem Statement

Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty}\left(\frac{2 n-1}{2 n+1}\right)^{n+1}$
## Solution

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

## Problem Kuznetsov Limits $7-26$