Puzzled by $\lim\limits_{x \to - \infty} \sqrt{x^2+x}-x$ 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?
 A: 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 $+\infty$
A: The difficulty here is in "factoring out" the $x$.  What you really did there is this:
You claim that $x^2+x=x^2(1+\frac{1}{x})$, and that $\sqrt{x^2(1+\frac{1}{x})}=\sqrt{x^2}\sqrt{1+\frac{1}{x}}$. This is perfectly fine so far, as long as the latter square root is defined.  The problem is in the last step: you've claimed, from here, that $\sqrt{x^2}=x$.  And that's not true!
Remember: in general, $\sqrt{x^2}=\lvert x\rvert$, not $x$.  And if $x<0$, then $\lvert x\rvert=-x$, not $x$.
So, it should be
$$
\sqrt{x^2+x}=\sqrt{x^2}\sqrt{1+\frac{1}{x}}=\lvert x\rvert\sqrt{1+\frac{1}{x}}=-x\sqrt{1+\frac{1}{x}}.
$$
A: If $x$ is negative, then the square root must be manipulated differently; that is, we have
$$\sqrt{x^2 + x} = \sqrt{x^2} \sqrt{1 + \frac 1 x} = |x| \sqrt{1 + \frac 1 x}$$
But since $x < 0$, $|x| = -x$. This leads us to taking the limit of
$$\frac{x}{-|x|\sqrt{1 + \frac 1 x} + x} = \frac{1}{-\sqrt{1 + \frac 1 x} + 1}$$
Letting $x \to -\infty$, the denominator tends to zero.
A: Setting $\displaystyle-\frac1x=h\iff x=-\frac1h$
$$\lim_{x\to-\infty}\left(\sqrt{x^2+x}-x\right)=\lim_{h\to0^+}\left(\sqrt{\frac1{h^2}-\frac1h}+\frac1h\right)$$
$$=\lim_{h\to0^+}\frac{\sqrt{1-h}+1}h\text{ as }h>0$$
$$=\lim_{h\to0^+}\frac{(1-h)-1}{h(\sqrt{1-h}-1)} $$
$$=-\lim_{h\to0^+}\frac1{\sqrt{1-h}-1}\text{ as } h\ne0\text{ as }h\to0^+ $$
$$=-\frac1{1-1}$$
A: $\newcommand{\+}{^{\dagger}}%
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 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
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 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
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 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
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 \newcommand{\pars}[1]{\left( #1 \right)}%
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 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
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 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{\mbox{Use always}\ \root{x^{2}} = \verts{x}\quad \mbox{and}\quad \verts{x} = x\,\sgn\pars{x}\ !}$
$$
{x \over \root{x^{2} + x} + x}
=
{\verts{x} \over \sgn\pars{x}\root{x^{2} + x} + \verts{x}}
=
{1 \over \sgn\pars{x}\root{1 + 1/x} + 1}\,,\qquad x \not= 0
$$

\begin{align}
&\lim_{x \to -\infty}{1 \over \sgn\pars{x}\root{1 + 1/x} + 1}
=
\lim_{x \to -\infty}{1 \over\ -\root{1 + 1/x} + 1} = +\infty
\\[3mm]&
\lim_{x \to +\infty}{1 \over \sgn\pars{x}\root{1 + 1/x} + 1}
=
\lim_{x \to +\infty}{1 \over\ +\root{1 + 1/x} + 1} = \half
\end{align}

Esentially, your error was the division $\ds{\root{x^{2} + x} \over x}$. It's, indeed:
$$
{\root{x^{2} + x} \over x} = -\,{\root{x^{2} + x} \over \verts{x}}\ \mbox{when}\
x < 0\quad\mbox{and}\quad 
{\root{x^{2} + x} \over x} = +\,{\root{x^{2} + x} \over \verts{x}}\ \mbox{when}\
x > 0
$$
