Limit of difference of two square roots

I need to find the limit, not sure what to do. $\lim_{x \to \infty} \sqrt{x^2 +ax} - \sqrt{x^2 +bx}$

I am pretty sure I have to divide by the largest degree which is x^2 but that gets me some weird numbers that don't seem to help.

-
Did you try multiplying by $\displaystyle\frac{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}$? –  t.b. Sep 6 '11 at 22:33
@Jordan: This is precisely the same type of question that you asked 51 minutes ago. Did you try to apply the same techniques? How long have you thought about this problem? What have you tried? What are the "weird numbers that don't seem to help"? –  JavaMan Sep 6 '11 at 22:33
I multiplied by the conjugate and I got (1+a/x) - (1+b/x) and I dont think that helps. –  user138246 Sep 6 '11 at 22:36
@Jordan If by that you mean the last expression in Edvin's answer, then that does help. $\lim_{x \to \infty} a/x$ is about as good as it gets. –  Dylan Moreland Sep 6 '11 at 22:54

Applying the formula $x^2-y^2=(x-y)(x+y)$ we get
$\sqrt{x^2+ax}-\sqrt{x^2+bx}= \frac{x^2+ax-x^2-bx}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}=\frac{x(a-b)}{x\left(\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}\right) }=\frac{a-b}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}}$
You could use the binomial expansion: write $\sqrt{x^2+ax}$ as $x\sqrt{1+{a\over x}}$, expand the square root out as $1+{a\over 2x}-{a^2\over 8x^2}+\ldots$, multiply back through by $x$ and subtract the corresponding calculation for the $\sqrt{x^2+bx}$ term. –  Steven Stadnicki Sep 6 '11 at 22:46