Find the value of $\lim_{x \to - \infty} \left( \sqrt{x^2 + 2x} - \sqrt{x^2 - 2x} \right)$ I am stuck on this.  I would like the algebraic explanation or trick(s) that shows that the equation below has limit of $-2$ (per the book). The wmaxima code of the equation below.
$$
\lim_{x \to - \infty} \left( \sqrt{x^2 + 2x} - \sqrt{x^2 - 2x} \right)
$$
I've tried factoring out an $x$ using the $\sqrt{x^2} = |x|$ trick.  That doesn't seem to work.  I get $1 - 1 = 0$ for the other factor meaning the limit is zero...but that's obviously not correct way to go about it :(
Thanks.

For $x>0$: For brevity let $A=\sqrt {x^2+2 x}.$ We have $(x+1)^2=A^2 +1>A^2>0$ so $x+1>A>0 . $ ..... So we have $$0<x+1-A=$$ $$=(x+1-A)\frac {x+1+A}{x+1+A}=\frac {(x+1)^2-A^2}{x+1+A}=\frac {1}{x+1+A}<1/x.$$ Therefore $$(i)\quad \lim_{x\to \infty} (x+1-A)=0.$$ For $x>2$: For brevity let $B=\sqrt {x^2-2 x}.$ We have $(x-1)^2=B^2+1>B^2>0$ so $x-1>B>0 . $ ..... So we have $$0<x-1-B=$$ $$=(x-1-B)\frac {x-1+B}{x-1+B}=\frac {(x-1)^2-B^2}{x-1+B}=\frac {1}{x-1+B}<1/(x-1).$$ Therefore $$(ii)\quad\lim_{x\to \infty}(x-1-B)=0.$$
We have $\sqrt {x^2+2 x}-\sqrt {x^2-2 x}=$ $A-B=(A-(x+1))-(B-(x-1))+2. $  From $(i)$ and $(ii)$ we have $$\lim_{x\to \infty} A-B=\lim_{x\to \infty}(A-(x+1))-(B-(x-1))+2=0+0+2=2.$$ 
The idea is that when $x$ is large, $A$ is close to $x+1$ and $B$ is close to $x-1$ so $A-B$ is close to $(x+1)-(x-1)=2.$
