Задача №2119

\[ \lim_{n\to\infty}\frac{(n+2)^2-(n-2)^2}{(n+3)^2} \lim_{n\to\infty}\frac{n^2+4n+4-\left(n^2-4n+4\right)}{(n+3)^2}=\\ =\lim_{n\to\infty}\frac{8n}{(n+3)^2} =\left[\frac{\infty}{\infty}\right] =\lim_{n\to\infty}\frac{\frac{8n}{n^2}}{\frac{(n+3)^2}{n^2}} =\lim_{n\to\infty}\frac{\frac{8}{n}}{\left(1+\frac{3}{n}\right)^2} =\frac{0}{1} =0 \]