# limit $\lim_{x\to+\infty }\sqrt{x^{2}+x}-x=\frac{1}{2}$ difference between $+\infty$ $-\infty$

someone can clarify a doubt risen solving a limit, please. I know $$\lim_{x\to+\infty }\sqrt{x^{2}+x}-x=\frac{1}{2}$$ $$\lim_{x\to-\infty }\sqrt{x^{2}+x}-x=-\infty$$ Solving the first one

$$\lim_{x\to+\infty }\sqrt{x^{2}+x}-x\cdot\frac{\sqrt{x^{2}+x}+x}{\sqrt{x^{2}+x}+x}$$ $$\lim_{x\to+\infty }\frac{x^{2}+x-x^{2}}{\sqrt{x^{2}+x}+x}$$ $$\lim_{x\to+\infty }\frac{1}{\sqrt{1+\frac{1}{x}}+1}=\frac{1}{2}$$ However doing the same steps,algebraically allowed, I obtein $$\lim_{x\to-\infty }\frac{1}{\sqrt{1+\frac{1}{x}}+1}=\frac{1}{2}$$ WRONG Where is the mistake? In the second case there is some step unallowed? Thank you so much for your help.

HINT:

$$\sqrt{x^2+x}=|x|\sqrt{1+\frac1x}\ne x\sqrt{1+\frac1x}$$

when $$x<0$$

SPOILER ALERT: Scroll over the highlighted area to reveal the solution.

Note that we have \begin{align}\lim_{x\to -\infty}\frac{x}{\sqrt{x^2+x}+x}&=\lim_{x\to -\infty}\frac{x}{|x|\sqrt{1+\frac1x}+x}\\\\&=\lim_{x\to \infty}\frac{x}{|x|\left(1+\frac{1}{2x}+O\left(\frac1{x^2}\right)\right)+x}\\\\&=\lim_{x\to -\infty}\frac{2x^2}{|x|+O\left(1\right)}\\\\&=+\infty\end{align}

$\sqrt {x ^2} = |{x}| \neq x$

• sorry, when I started writing this your comment was not yet there. I agree that there is no difference, yours is better because it is more detailed. should I delete my comment? – Paul Aug 16 '16 at 15:52
• Paul, no worry. That has happened to me also (i.e., I've posted a solution without noticing that a similar one was previously posted). -Mark – Mark Viola Aug 17 '16 at 17:40