Question

\(\lim\limits_{x\rightarrow+\infty}\dfrac{2x\left(\sqrt{4x^2-2x}+\sqrt[3]{3x^2-8x^3}\right)}{5x-1}\)

Answer 1

Ko viết lim cho lẹ nhé, tự hiểu

\(=\dfrac{2x\left(\sqrt{4x^2-2x}-2x+2x+\sqrt[3]{3x^2-8x^3}\right)}{5x-1}\)

\(=\dfrac{2x\left(\dfrac{-2x}{\sqrt[]{4x^2-2x}+2x}+\dfrac{3x^2}{4x^2-2x\sqrt[3]{3x^2-8x^3}+\sqrt[3]{\left(3x^2-8x^3\right)^2}}\right)}{5x-1}\)

\(=\dfrac{2\left(\dfrac{-2}{\sqrt[]{4-\dfrac{2}{x}}+2}+\dfrac{3}{4-2.\sqrt[3]{\dfrac{3}{x}-8}+\sqrt[3]{\left(\dfrac{3}{x}-8\right)^2}}\right)}{5-\dfrac{1}{x}}\)

\(=\dfrac{2\left(\dfrac{-2}{2+2}+\dfrac{3}{4-2.\left(-2\right)+2}\right)}{5}\)