## Problem Statement

Calculate the limit of the numerical sequence:

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

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

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

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

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

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

Problem Kuznetsov Limits 7-1