Giá trị của giới hạn 

- Hướng dẫn giải

$\begin{array}{l}\underset{x\to +\infty }{\lim }\left(\sqrt{x^2+x}-\sqrt[3]{x^3-x^2}\right)\\ =\underset{x\to +\infty }{\lim }\left(\sqrt{x^2+x}-x+x-\sqrt[3]{x^3-x^2}\right)\\ =\underset{x\to +\infty }{\lim }\left(\frac{x}{\sqrt{x^2+x}+x}+\frac{x^2}{x^2+x\sqrt[3]{x^3-x^2}+\sqrt[3]{{\left(x^3-x^2\right)}^2}}\right)\\ =\underset{x\to +\infty }{\lim }\left(\frac{1}{\sqrt{1+\text{}\frac{1}{x}}+1\text{\hspace{0.17em}\hspace{0.17em}}}+\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{1+\text{}\sqrt[3]{1-\frac{1}{x}}+\text{}\sqrt[3]{{(1-\frac{1}{x})}^2}\text{\hspace{0.17em}\hspace{0.17em}}}\right)\\ =\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\end{array}$

Đáp án cần chọn là: A