Задача №2110

\[ \lim_{n\to\infty}\frac{(2n+1)^3-(2n+3)^3}{(2n+1)^2+(2n+3)^2} =\lim_{n\to\infty}\frac{8n^3+12n^2+6n+1-\left(8n^3+36n^2+54n+27\right)}{(2n+1)^2+(2n+3)^2}=\\ =\lim_{n\to\infty}\frac{-24n^2-48n-26}{(2n+1)^2+(2n+3)^2} =\left[\frac{\infty}{\infty}\right] =\lim_{n\to\infty}\frac{\frac{-24n^2-48n-26}{n^2}}{\frac{(2n+1)^2+(2n+3)^2}{n^2}} =\lim_{n\to\infty}\frac{-24-\frac{48}{n}-\frac{26}{n^2}}{\left(2+\frac{1}{n}\right)^2+\left(2+\frac{3}{n}\right)^2} =\frac{-24}{8} =-3. \]