Evaluate the following limit without the use of L'Hopital's rule:

$\displaystyle \lim_{x\to0}\frac{sin^2x}{\sqrt{1+xsinx}-cosx} $

My first instinct was to multiply by the conjugate, leaving me with:

$\displaystyle \lim_{x\to0}\frac{sin^2x(\sqrt{1+xsinx}+cosx)}{1+x sinx-cos^2x} $

And from here I tried rearranging as best I could but could not end up with any form without zero in the denominator.