Задача №2111

\[ \lim_{n\to\infty}\frac{n^3-(n-1)^3}{(n+1)^4-n^4} =\lim_{n\to\infty}\frac{n^3-\left(n^3-3n^2+3n-1\right)}{n^4+4n^3+6n^2+4n+1-n^4}=\\ =\lim_{n\to\infty}\frac{3n^2-3n+1}{4n^3+6n^2+4n+1} =\left[\frac{\infty}{\infty}\right] =\lim_{n\to\infty}\frac{\frac{3n^2-3n+1}{n^3}}{\frac{4n^3+6n^2+4n+1}{n^3}} =\lim_{n\to\infty}\frac{\frac{3}{n}-\frac{3}{n^2}+\frac{1}{n^3}}{4+\frac{6}{n}+\frac{4}{n^2}+\frac{1}{n^3}} =\frac{0}{4} =0. \]