Question

\(\lim\limits_{x\rightarrow+\infty}\dfrac{2x-\sqrt{3x^2+2}}{5x+\sqrt{x^2+1}}\)

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

\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt[3]{1+x^4+x^6}}{\sqrt{1+x^3+x^4}}\)

Answer 1

1/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{2x}{x}-\sqrt{\dfrac{3x^2}{x^2}+\dfrac{2}{x^2}}}{\dfrac{5x}{x}+\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}}=\dfrac{2-\sqrt{3}}{5+1}=\dfrac{2-\sqrt{3}}{6}\)

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

3/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt[3]{\dfrac{x^6}{x^6}+\dfrac{x^4}{x^6}+\dfrac{1}{x^6}}}{\sqrt{\dfrac{x^4}{x^4}+\dfrac{x^3}{x^4}+\dfrac{1}{x^4}}}=-1\)