Question

Limita când n tinde la infinit din

Answer 1

$\lim\limits_{n\to \infty}n\Big[\sqrt[3]{n^3+3n^2+1}-(n+1)\Big] =\\\\=\lim\limits_{n\to \infty}n\Big[\sqrt[3]{n^3(1+\frac{3}{n}+\frac{1}{n^3}})-n(1+\frac{1}{n})\Big] = \\ \\ = \lim\limits_{n\to \infty}n\Big[n\sqrt[3]{(1+\frac{3}{n}+\frac{1}{n^3}})-n(1+\frac{1}{n})\Big] = \\ \\ = \lim\limits_{n\to \infty}n^2\Big[\sqrt[3]{1+\frac{3}{n}+\frac{1}{n^3}}-1-\frac{1}{n}\Big]= \\ \\ = \lim\limits_{n\to \infty}\dfrac{\sqrt[3]{1+\frac{3}{n}+\frac{1}{n^3}}-1-\frac{1}{n}}{\frac{1}{n^2}} =$

$= \lim\limits_{n\to \infty}\dfrac{(1+\frac{3}{n}+\frac{1}{n^3})^{\frac{1}{3}}-1-\frac{1}{n}}{\frac{1}{n^2}}=\\ \\ \dfrac{1}{n} = t \\ n\to \infty \Rightarrow t\to 0 \\\\ = \lim\limits_{t\to 0}\dfrac{(1+3t+t^3)^{\frac{1}{3}}-1-t}{t^2} \overset{\frac{0}{0}}{=}\\ \\ \overset{\frac{0}{0}}{=}\lim\limits_{t\to 0}\dfrac{\frac{1}{3}(1+3t+t^2)^{-\frac{2}{3}}(3+3t^2)-1}{2t} \overset{\frac{0}{0}}{=}$

$\overset{\frac{0}{0}}{=} \lim\limits_{t\to 0}\dfrac{-\frac{2}{9}(1+3t+t^2)^{-\frac{5}{3}}(3+3t^2)(3+3t^2)+\frac{1}{3}(1+3t+t^2)^{-\frac{2}{3}}(6t)-1}{2} =\\ \\ = \dfrac{-\frac{2}{9}\cdot 3\cdot 3}{2} = \boxed{-1}$