Question

[dung doan]

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

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

Answer 1

[Hoàng Tử Hà]

1/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{\dfrac{4x^2}{x^2}-\dfrac{2}{x^2}}+\sqrt[3]{\dfrac{x^3}{x^3}+\dfrac{1}{x^3}}}{-\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}-\dfrac{x}{x}}=\dfrac{-1+1}{-1-1}=0\)

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