TITLE: Puzzled by $\lim\limits_{x \to - \infty} \sqrt{x^2+x}-x$
QUESTION [6 upvotes]: I am preparing for the next Semester and therefore review a few of my Analysis I limits, I have found this example in C.T. Michaels Analysis I:

Compute $ \displaystyle \lim_{x \to - \infty} \sqrt{x^2+x}-x$

Aprior to this exercise I computed the same limit but as $x$ approaches $\infty$ rather than $- \infty$. So I thought that this should be a piece of cake, but apparently the $- \infty$ makes all the difference for me. 
My approach: This is the general approach I take when it comes to roots, especially square roots. Consider: $$ \sqrt{x^2+x}-x= \left(\sqrt{x^2+x}-x\right)\cdot \frac{\sqrt{x^2+x}+x}{\sqrt{x^2+x}+x}= \frac{x^2+x-x^2}{\sqrt{x^2+x}+x}=\frac{x}{\sqrt{x^2+x}+x}$$
Factoring out an $x$ will get me to: $$ \frac{x}{x\left(\sqrt{1+\frac{1}{x}}+1\right)}=\frac{1}{\sqrt{1+\frac{1}{x}}+1}$$
So as I take the limit of the above expressing as $x$ approaches $\infty$ I obtain the correct answer $1/2$.
However, when I study the limit as $x$ approaches $- \infty$ I don't see how that would make a difference since $1 / - \infty=0$, but the correct answer in that case would be $\infty$ 
http://www.wolframalpha.com/input/?i=lim+x+to+-infty+sqrt%28x%5E2%2Bx%29-x
My question(s):
Where is/are my mistakes?
Is it not possible to use the same methods for $- \infty$ as for $\infty$ when studying limits?

REPLY [2 votes]: In addition to all the other answers, just note that if we set 
$y = -x$ then we have
$$\lim_{y \to +\infty} \sqrt{y^2 - y} + y$$ which is clearly $+\infty$