Question

Найти пределы...... .

Answer 1

$\displaystyle \lim_{n \to \infty} \frac{n}{(n+3)^2}=\lim_{n \to \infty} \frac{n}{n^2+6n+9}=\\=\lim_{n \to \infty} \frac{n:n}{n^2:n+6n:n+9:n}=\lim_{n \to \infty} \frac{1}{n+6+\frac{9}{n}}=0;$

$\displaystyle \lim_{n \to \infty} \frac{2+10n}{5n-1}=\lim_{n \to \infty} \frac{2:5n+10n:5n}{5n:5n-1:5n}=\\\lim_{n \to \infty} \frac{\frac{2}{5n}+2}{1-\frac{1}{5n}}=\frac{2}{1}=2$

$\displaystyle \lim_{n \to \infty} \frac{4+n^3}{2n}=\lim_{n \to \infty} \frac{4:n+n^3:n}{2n:n}=\\\lim_{n \to \infty} \frac{\frac{4}{n}+n^2}{2}=\infty$