Question

Tính\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt[3]{n^3+1}-\sqrt[3]{n^3+2}\right)\)

Answer 1

Sửa đề: \(\lim\limits_{n\rightarrow+\infty}\left(\sqrt[3]{n^3+1}-\sqrt[3]{n^3+2}\right)\)

\(=\lim\limits_{n\rightarrow+\infty}\dfrac{n^3+1-n^3-2}{\sqrt[3]{\left(n^3+1\right)^2}+\sqrt[3]{\left(n^3+1\right)\left(n^3+2\right)}+\sqrt[3]{\left(n^3+2\right)^2}}\)

\(=\lim\limits_{n\rightarrow+\infty}\dfrac{-1}{\sqrt[3]{\left(n^3+1\right)^2}+\sqrt[3]{\left(n^3+1\right)\left(n^3+2\right)}+\sqrt[3]{\left(n^3+2\right)^2}}\)

\(=\lim\limits_{n\rightarrow+\infty}\dfrac{-\dfrac{1}{n^2}}{\sqrt[3]{\left(1+\dfrac{1}{n}\right)^2}+\sqrt[3]{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{2}{n}\right)}+\sqrt[3]{\left(1+\dfrac{2}{n}\right)^2}}\)

\(=0\)