Граница на функция3

[emil3993]

Здравейте математици! Може ли решението? Благодаря предварително.
[tex]\lim_{n \to \infty}\sqrt[3]{1-n^3}+n[/tex]

[Добромир Глухаров]

$(a+b)(a^2-ab+b^2)=a^3+b^3$

$\lim_{n\to\infty}\sqrt[3]{1-n^3}+n=\lim_{n\to\infty}\frac{(\sqrt[3]{1-n^3}+n)(\sqrt[3]{(1-n^3)^2}-n\sqrt[3]{1-n^3}+n^2)}{\sqrt[3]{(1-n^3)^2}-n\sqrt[3]{1-n^3}+n^2}=\\=\lim_{n\to\infty}\frac{1-n^3+n^3}{n^2\sqrt[3]{(\frac{1}{n^3}-1)^2}-n^2\sqrt[3]{\frac{1}{n^3}-1}+n^2}=\\=\lim_{n\to\infty}\frac{1}{n^2}\cdot\frac{1}{1-(-1)+1}=0\cdot\frac{1}{3}=0$

[Knowledge Greedy]

[tex]\lim_{n \to \infty} \left ( \sqrt[3]{1-n^3}+n \right )=\lim_{n \to \infty} \frac{1-\cancel{n^3}+\cancel{n^3}}{ (\sqrt[3]{1-n^3})^2-\sqrt[3]{1-n^3}n+n^2} \sim \lim_{n \to \infty} \frac{1}{3n^2}=0[/tex]

[emil3993]

Благодаря.