## Problem Statement

Calculate the limit of the numerical sequence:

$$
\lim _{n \rightarrow \infty} \frac{(n+4)!-(n+2)!}{(n+3)!}
$$
## Solution

$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(n+4)!-(n+2)!}{(n+3)!}=\lim _{n \rightarrow \infty}\left(\frac{(n+4)!}{(n+3)!}-\frac{(n+2)!}{(n+3)!}\right)= \\
& =\lim _{n \rightarrow \infty}\left((n+4)-\frac{1}{n+3}\right)=\lim _{n \rightarrow \infty}\left(\frac{(n+4)(n+3)}{n+3}-\frac{1}{n+3}\right)= \\
& =\lim _{n \rightarrow \infty} \frac{n^{2}+4 n+3 n+12-1}{n+3}=\lim _{n \rightarrow \infty} \frac{n^{2}+7 n+11}{n+3}= \\
& =\lim _{n \rightarrow \infty} \frac{\frac{1}{n}\left(n^{2}+7 n+11\right)}{\frac{1}{n}(n+3)}=\lim _{n \rightarrow \infty} \frac{n+7+\frac{11}{n}}{1+\frac{3}{n}}= \\
& =\left\{\frac{\infty+7+0}{1+0}=\frac{\infty}{1}\right\}=\infty
\end{aligned}
$$

## Problem Kuznetsov Limits 6-9