Question

Mọi người giải giúp em bài này với!

1,\(\lim\limits_{x\rightarrow+\infty}\frac{x^4+8x}{x^3+2x^2+x+2}\)

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

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

Answer 1

\(a=\lim\limits_{x\rightarrow+\infty}\frac{x+\frac{8}{x^2}}{1+\frac{2}{x}+\frac{1}{x^2}+\frac{2}{x^3}}=\frac{+\infty}{1}=+\infty\)

\(b=\lim\limits_{x\rightarrow-\infty}\frac{x\left(\frac{1}{x}+3\right)}{\left|x\right|\sqrt{2+\frac{3}{x^2}}}=\lim\limits_{x\rightarrow-\infty}\frac{x\left(\frac{1}{x}+3\right)}{-x\sqrt{2+\frac{3}{x^2}}}=\frac{3}{-\sqrt{2}}=\frac{-3\sqrt{2}}{2}\)

\(c=\lim\limits_{x\rightarrow-\infty}\frac{x^2\sqrt[3]{\frac{1}{x^6}+\frac{1}{x^2}+1}}{x^2\sqrt{\frac{1}{x^2}+\frac{1}{x}+1}}=\frac{1}{1}=1\)