Tính I= lim x đến dương vô cùng căn x^3+2x^2- cưn x^2-2x

Đáp án đúng là: A

$$\begin{gathered} I=\underset{x\to +\infty }{\lim }\left(\sqrt[3]{x^3+2x^2}-\sqrt{x^2-2x}\right) \\ =\underset{x\to +\infty }{\lim }\left[\left(\sqrt[3]{x^3+2x^2}-x+1\right)+\left(x-1-\sqrt{x^2-2x}\right)\right] \\ =\underset{x\to +\infty }{\lim }\frac{\left(\sqrt[3]{x^3+2x^2}-x+1\right)\left[{\sqrt[3]{x^3+2x^2}}^2+\sqrt[3]{x^3+2x^2}\left(x-1\right)+{\left(x-1\right)}^2\right]}{{\sqrt[3]{x^3+2x^2}}^2+\sqrt[3]{x^3+2x^2}\left(x-1\right)+{\left(x-1\right)}^2} \\ +\underset{x\to +\infty }{\lim }\frac{\left(x-1-\sqrt{x^2-2x}\right)\left(x-1+\sqrt{x^2-2x}\right)}{x-1+\sqrt{x^2-2x}} \\ =\underset{x\to +\infty }{\lim }\frac{{\displaystyle \left(x^3+2x^2\right)-{\left(x-1\right)}^3}}{{\displaystyle {\sqrt[3]{x^3+2x^2}}^2+\sqrt[3]{x^3+2x^2}\left(x-1\right)+{\left(x-1\right)}^2}}+\underset{x\to +\infty }{\lim }\frac{{\displaystyle {\left(x-1\right)}^2-\left(x^2-2x\right)}}{{\displaystyle x-1+\sqrt{x^2-2x}}} \\ =\underset{x\to +\infty }{\lim }\frac{{\displaystyle 5x^2-3x+1}}{{\displaystyle {\sqrt[3]{x^3+2x^2}}^2+\sqrt[3]{x^3+2x^2}\left(x-1\right)+{\left(x-1\right)}^2}}+\underset{x\to +\infty }{\lim }\frac{{\displaystyle 1}}{{\displaystyle x-1+\sqrt{x^2-2x}}} \\ =\frac{{\displaystyle 5}}{{\displaystyle 1+1+1}}+0=\frac{{\displaystyle 5}}{{\displaystyle 3}}. \end{gathered}$$