## Problem Statement

Calculate the limit of the numerical sequence:

$\lim _{n \rightarrow \infty}\left(\frac{7 n^{2}+18 n-15}{7 n^{2}+11 n+15}\right)^{n+2}$
## Solution

$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{7 n^{2}+18 n-15}{7 n^{2}+11 n+15}\right)^{n+2}= \\
& =\lim _{n \rightarrow \infty}\left(\frac{7 n^{2}+11 n+15+7 n-30}{7 n^{2}+11 n+15}\right)^{n+2}=
\end{aligned}
$$

$$
\begin{aligned}
& =\lim _{n \rightarrow \infty}\left(1+\frac{7 n-30}{7 n^{2}+11 n+15}\right)^{n+2}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{7 n^{2}+11 n+15}{7 n-30}\right)}\right)^{n+2}= \\
& =\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{7 n^{2}+11 n+15}{7 n-30}\right)}\right)^{\left(\frac{7 n^{2}+11 n+15}{7 n-30}\right)\left(\frac{7 n-30}{7 n^{2}+11 n+15}\right)(n+2)}= \\
& =\left(\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{7 n^{2}+11 n+15}{7 n-30}\right)}\right)^{\left(\frac{7 n^{2}+11 n+15}{7 n-30}\right)}\right)_{n \rightarrow \infty}^{\lim _{n \rightarrow \infty}\left(\frac{7 n-30}{7 n^{2}+11 n+15}\right)(n+2)}=
\end{aligned}
$$

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

$=e^{\lim _{n \rightarrow \infty}\left(\frac{7 n-30}{7 n^{2}+11 n+15}\right)(n+2)}=e^{\lim _{n \rightarrow \infty} \frac{(7 n-30)(n+2)}{7 n^{2}+11 n+15}}=$

$=e^{\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}(7 n-30)(n+2)}{n^{2}\left(7 n^{2}+11 n+15\right)}}=e^{\lim _{n \rightarrow \infty} \frac{\left(7-\frac{30}{n}\right)\left(1+\frac{1}{n}\right)}{7+\frac{11}{n}+\frac{15}{n^{2}}}}=$

$=e^{\frac{(7-0)(1+0)}{7+0+0}}=e$

## Problem Kuznetsov Limits 7-25