## Problem Statement

Calculate the limit of the numerical sequence:

$\lim _{n \rightarrow \infty}\left(\frac{3 n^{2}-5 n}{3 n^{2}-5 n+7}\right)^{n+1}$
## Solution

$\lim _{n \rightarrow \infty}\left(\frac{3 n^{2}-5 n}{3 n^{2}-5 n+7}\right)^{n+1}=\lim _{n \rightarrow \infty}\left(\frac{3 n^{2}-5 n+7}{3 n^{2}-5 n}\right)^{-n-1}=$

$=\lim _{n \rightarrow \infty}\left(1+\frac{7}{3 n^{2}-5 n}\right)^{-n-1}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{3 n^{2}-5 n}{7}\right)}\right)^{-n-1}=$

$=\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{3 n^{2}-5 n}{7}\right)}\right)^{\left(\frac{3 n^{2}-5 n}{7}\right)\left(\frac{7}{3 n^{2}-5 n}\right)(-n-1)}=$

![](https://cdn.mathpix.com/cropped/2024_05_22_ae19746d5d2a92572015g-04.jpg?height=263&width=1223&top_left_y=1148&top_left_x=166)

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

$=e^{\lim _{n \rightarrow \infty}\left(\frac{7}{3 n^{2}-5 n}\right)(-n-1)}=e^{\lim _{n \rightarrow \infty} \frac{7(-n-1)}{3 n^{2}-5 n}}=$

$=e^{\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}} 7(-n-1)}{n^{2}\left(3 n^{2}-5 n\right)}}=e^{\lim _{n \rightarrow \infty} \frac{7\left(-\frac{1}{n}-\frac{1}{n^{2}}\right)}{3-\frac{5}{n^{2}}}}=e^{\frac{7(-0-0)}{3-0}}=e^{0}=1$

## Problem Kuznetsov Limits 7-21