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!! 
 A: $$\lim _{ x\rightarrow y } \frac { \sin ^{ 2 } x-\sin ^{ 2 } y }{ x^{ 2 }-y^{ 2 } } =\lim _{ x\rightarrow y } \frac { \left( \sin { x-\sin { y }  }  \right) \left( \sin { x+\sin { y }  }  \right)  }{ \left( x-y \right) \left( x+y \right)  } =\\ =\lim _{ x\rightarrow y } \frac { \left( 2\sin { \frac { x-y }{ 2 }  } \cos { \frac { x+y }{ 2 }  }  \right) \left( \sin { x+\sin { y }  }  \right)  }{ \left( x-y \right) \left( x+y \right)  } =\\ =\lim _{ x\rightarrow y } \frac { \sin { \frac { x-y }{ 2 }  }  }{ \frac { \left( x-y \right)  }{ 2 }  } \frac { \cos { \frac { x+y }{ 2 }  } \left( \sin { x+\sin { y }  }  \right)  }{ \left( x+y \right)  } =\\ =\cos { y } \frac { 2\sin { y }  }{ 2y } =\frac { \sin { y\cos { y }  }  }{ y } $$
A: Take $x=y+h$
$$\lim_{h\to0}{\sin^2(y+h)-\sin^2y\over (y+h)^2-y^2}=\lim_{h\to0}{\sin(y+h)-\sin(y)\over h}\cdot{\sin(y+h)+\sin(y)\over 2y+h}\\=\cos(y)\cdot{2\sin(y)\over 2y}=\cos(y)\cdot{\sin(y)\over y}$$
A: $\lim_\limits{x\to y} \frac{\sin^2 x - \sin^2 y}{x^2-y^2} =\lim_\limits{x\to y} \left(\frac{\sin x - \sin y}{x-y}\right)\left(\frac{\sin x + \sin y}{x+y}\right)\\
(\frac {d}{dy} \sin y)(\frac {\sin y}{y})$
A: $$\begin{align*}\lim_{x\to y} \frac{\sin^2(x)-\sin^2(y)}{x^2-y^2}&=\lim_{x\to y}\frac{\sin x-\sin y}{x-y}\frac{\sin x+\sin y}{x+y}\\&=\frac{2\sin y}{2y}\lim_{x\to y}\frac{\sin x-\sin y}{x-y}\\&=\frac{\sin y}{y}(\sin y)'\\&=\frac{\sin y \cos y}{y}\end{align*}$$
$$=\frac{\sin2y}{2y}$$
A: Use the following formulas
$$\begin{align}\sin{p}+\sin{q}=&2\sin{p+q\over 2}\cos{p-q\over 2}\\
\sin{p}-\sin{q}=&2\sin{p-q\over 2}\cos{p+q\over 2}
\end{align}$$
to rewrite
$${\sin^2{x}-\sin^2{y}\over x^2-y^2}=2{\sin{x-y\over 2}\over{x-y\over 2}}{\sin{x+y\over 2}\over x+y}\cos{x+y\over 2}\cos{x-y\over 2}$$
And so the limit is 
$${\sin{y}\cos{y}\over y}$$
Because ${\sin{x-y\over 2}\over {x-y\over 2}}\to 1$ (the rest is straightforward)
A: $\lim\limits_{x\rightarrow y}\frac{\sin^2x-\sin^2y}{x^2-y^2} = \lim\limits_{x\rightarrow y}\frac{\sin^2x-\sin^2y}{x-y}\frac{1}{x+y} = \left(\frac{d}{d y}\sin^2y\right)\frac{1}{y+y} = \frac{2\sin y\cos y}{2y} = \frac{\sin y\cos y}{y}$
It can also be written a few other ways, as
$\frac{\sin y \cos y}{y} = \frac{\sin2y}{2y} = \text{sinc} \ {2y}$
