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.

So:
$$
\lim_{x\to-\infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 2x}) = \lim_{x\to-\infty}(\frac{x^2 + 2x - x^2 + 2x}{\sqrt{x^2 + 2x} + \sqrt{x^2 - 2x}}) = \lim_{x\to-\infty}(\frac{4x}{\sqrt{x^2 + 2x} + \sqrt{x^2 - 2x}}) = \lim_{x\to-\infty}(\frac{-4}{\sqrt{1 + \frac{2}{x}} + \sqrt{1 - \frac{2}{x}}}) = -2
$$
