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.

Applying the formula $x^2-y^2=(x-y)(x+y)$ we get
$\sqrt{x^2+ax}-\sqrt{x^2+bx}= \frac{x^2+ax-x^2-bx}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}=\frac{x(a-b)}{x\left(\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}\right) }=\frac{a-b}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}}$ 
now you can take the limit.
