Question

[Phạm Minh Khôi]

lim x tiến tới vô cùng (\(\sqrt{x^2+x}\)-\(\sqrt[3]{x^3-x^2}\))

Answer 1

[Hồng Phúc]

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

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

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

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

\(=\dfrac{1}{\sqrt{1+0}+1}+\dfrac{1}{1+\sqrt[3]{1-0}+\sqrt[3]{\left(1-0\right)^2}}\)

\(=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)