## Calculus: Early Transcendentals 8th Edition

The asymptotes are 1) Line $y=-1$ is the horizontal asymptote when $x\to-\infty$; 2) Line $y=1$ is the horizontal asymptote when $x\to+\infty$.
From the graph we suspect that 1) Line $y=-1$ is the horizontal asymptote when $x\to-\infty$; 2) Line $y=1$ is the horizontal asymptote when $x\to+\infty$. Let us prove this. 1) $$\lim_{x\to-\infty}(\sqrt{x^2+x+1}-\sqrt{x^2-x})=\lim_{x\to-\infty}(\sqrt{x^2+x+1}-\sqrt{x^2-x})\cdot\frac{\sqrt{x^2+x+1}+\sqrt{x^2-x}}{\sqrt{x^2+x+1}+\sqrt{x^2-x}}=\lim_{x\to-\infty}\frac{\sqrt{x^2+x+1}^2-\sqrt{x^2-x}^2}{\sqrt{x^2+x+1}+\sqrt{x^2-x}}=\lim_{x\to-\infty}\frac{x^2+x+1-x^2+x}{\sqrt{x^2+x+1}+\sqrt{x^2-x}}=\lim_{x\to-\infty}\frac{2x+1}{\sqrt{x^2+x+1}+\sqrt{x^2-x}}=\lim_{x\to-\infty}\frac{|x|\left(-2+\frac{1}{|x|}\right)}{|x|\left(\sqrt{1-\frac{1}{|x|}+\frac{1}{|x|^2}}+\sqrt{1+\frac{1}{|x|}}\right)}=\lim_{x\to-\infty}\frac{\left(-2+\frac{1}{|x|}\right)}{\left(\sqrt{1-\frac{1}{|x|}+\frac{1}{|x|^2}}+\sqrt{1+\frac{1}{|x|}}\right)}=\left[\frac{-2+0}{\sqrt{1-0+0}+\sqrt{1+0}}\right]=-1,$$ which we needed to show. 2) $$\lim_{x\to+\infty}(\sqrt{x^2+x+1}-\sqrt{x^2-x})=\lim_{x\to+\infty}(\sqrt{x^2+x+1}-\sqrt{x^2-x})\cdot\frac{\sqrt{x^2+x+1}+\sqrt{x^2-x}}{\sqrt{x^2+x+1}+\sqrt{x^2-x}}=\lim_{x\to+\infty}\frac{\sqrt{x^2+x+1}^2-\sqrt{x^2-x}^2}{\sqrt{x^2+x+1}+\sqrt{x^2-x}}=\lim_{x\to+\infty}\frac{x^2+x+1-x^2+x}{\sqrt{x^2+x+1}+\sqrt{x^2-x}}=\lim_{x\to+\infty}\frac{2x+1}{\sqrt{x^2+x+1}+\sqrt{x^2-x}}=\lim_{x\to+\infty}\frac{x\left(2+\frac{1}{x}\right)}{x\left(\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+\sqrt{1-\frac{1}{x}}\right)}=\lim_{x\to+\infty}\frac{\left(2+\frac{1}{x}\right)}{\left(\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+\sqrt{1-\frac{1}{|x}}\right)}=\left[\frac{2+0}{\sqrt{1+0+0}+\sqrt{1-0}}\right]=1,$$ which we needed to show.