Question

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

Answer 1

Th1: \(\lim\limits_{x\rightarrow+\infty}\dfrac{x}{\sqrt{x^2+x+1}}\)

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

\(=\dfrac{1}{\sqrt{1+0+0}}=\dfrac{1}{1}=1\)

TH2: \(\lim\limits_{x\rightarrow-\infty}\dfrac{x}{\sqrt{x^2+x+1}}\)

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

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