## Problem Statement

Calculate the limit of the numerical sequence:

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

$\lim _{n \rightarrow \infty} \frac{(3-4 n)^{2}}{(n-3)^{3}-(n+3)^{3}}=$

$\lim _{n \rightarrow x} \frac{9-24 n+16 n^{2}}{n l^{3}-3 \cdot n l^{2} \cdot 3+3 \cdot n \cdot 3^{2}-3^{3}-n^{3}-3 \cdot n^{2} \cdot 3-3 \cdot n \cdot 3^{2}-3^{3}}=$

$=\lim _{n \rightarrow x} \frac{9-\cdot 24 n+16 n^{2}}{-3 \cdot n^{2} \cdot 3-3^{3}-3 \cdot n^{2} \cdot 3-3^{3}}=\lim _{n \rightarrow \infty} \frac{9-24 n+16 n^{2}}{-2 \cdot 3^{2}\left(n^{2}+3\right)}=$

$=\lim _{n \rightarrow x} \frac{\frac{1}{n^{2}}\left(9-24 n+16 n^{2}\right)}{\frac{1}{n^{2}}(-2) \cdot 3^{2}\left(n n^{2}+3\right)}=\lim _{n \rightarrow x} \frac{\frac{g}{n^{2}}-\frac{24}{n}+16}{-2 \cdot 3^{2}\left(1+\frac{5}{n^{2}}\right)}=$

$=\frac{0-0+16}{-2 \cdot 3^{2}(1+0)}=\frac{16}{-2 \cdot 3^{2}}=-\frac{8}{9}$

## Problem Kuznetsov Limits 2-8(2)