Evaluating $\displaystyle\lim_{x\rightarrow y}\frac{\sin^2x-\sin^2y}{x^2-y^2}$ 
I have to evaluate $\displaystyle\lim_{x\rightarrow y}\frac{\sin^2x-\sin^2y}{x^2-y^2}$

I tried several ways like replacing $x^2-y^2$ by $z$ and then solving and breaking the denominator into $(x+y)(x-y)$ and then replacing $(x-y)$ by $z$ and then solving but always I end up with complicated things which are nowhere near the answer.
How do I solve?
Thanks for any help!! 

$\lim\limits_{x\rightarrow y}\frac{\sin^2x-\sin^2y}{x^2-y^2} = \lim\limits_{x\rightarrow y}\frac{\sin^2x-\sin^2y}{x-y}\frac{1}{x+y} = \left(\frac{d}{d y}\sin^2y\right)\frac{1}{y+y} = \frac{2\sin y\cos y}{2y} = \frac{\sin y\cos y}{y}$
It can also be written a few other ways, as
$\frac{\sin y \cos y}{y} = \frac{\sin2y}{2y} = \text{sinc} \ {2y}$
