Question

срочно"'!!!пожалуйста

Answer 1

$\lim_{n \to \infty} \frac{n^3+5n^2}{3n^3-n^2-n} = \lim_{n \to \infty} \frac{n^3(1+\frac{5}{n})}{n^3(3-\frac{1}{n}-\frac{1}{n^2})} = \lim_{n \to \infty} \frac{1+\frac{5}{n}}{3-\frac{1}{n}-\frac{1}{n^2}} = \\ =\frac{1+0}{3-0-0} = \frac{1}{3}$

Answer 2

$lim_{n\to \infty}\frac{n^3+5n^2}{3n^3-n^2-n}=[\frac{:n^3}{:n^3}]=lim_{n\to \infty}\frac{1+\frac{5}{n}}{3-\frac{1}{n}-\frac{1}{n^2}}=\frac{1+0}{3-0-0}=\frac{1}{3}$