Question

Решите номер 5.Есть вложение.

Answer 1

$\lim\limits _{n \to \infty}\frac{(\sqrt{n^2+1}+n)^2}{\sqrt[3]{27n^6+1}}=\Big [\, \frac{:n^2}{:n^2}\, \Big ]=\lim\limits _{n \to \infty}\frac{(\frac{\sqrt{n^2+1}}{n}+\frac{n}{n})^2}{\frac{\sqrt[3]{27n^6+1}}{n^2}}=\lim\limits _{n \to \infty}\frac{(\sqrt{1+\frac{1}{n^2}}+1)^2}{\sqrt[3]{27+\frac{1}{n^6}}}=\\\\=\frac{(1+1)^2}{\sqrt[3]{27}}=\frac{4}{3}$