Question

Answer 1

\(lim\left(\sqrt{x^2-x+1}\right)-x\)

\(limx\left[\left(\sqrt{1-\dfrac{1}{x}+\dfrac{1}{x^2}}\right)-1\right]\)

\(=x\left(\sqrt{1}-1\right)=0\)

Answer 2

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

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