Use Cauchy's Theorem to find $\lim_{(x,y)\to(0,0)} \frac{\cos(x) - \cos(y)}{x^2 - y^2}$. I'm starting with Cauchy's theorem and I have this exercise... I don't understand how the theorem would help me in finding the limit.

$$f(x,y)=\left\{
\begin{array}{ll}
      \lim_{(x,y)\to(0,0)} \frac{\cos(x) - \cos(y)}{x^2 - y^2} & x \neq y \\
      -\frac{1}{2} &  x=y\\
\end{array} 
\right.$$

Suggestions? Perhaps I just need to see how what the theorem states is applicable to this situation... thanks!
 A: HINT:  Without The Mean Value Theorem:
Use the Prosthaphaeresis Formula
$$\cos(x)-\cos(y)=-2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)$$
Then write 
$$\frac{\cos(x)-\cos(y)}{x^2-y^2}=-2\left(\frac{\sin\left(\frac{x+y}{2}\right)}{x+y}\right)\left(\frac{\sin\left(\frac{x-y}{2}\right)}{x-y}\right)$$

HINT:  With The Mean Value Theorem:
If $x<y$, there exists a number $\xi\in (x,y)$ such that 
$$\frac{\cos(x)-\cos(y)}{x^2-y^2}=-\frac{\sin(\xi)}{2\xi}$$
If $y<x$, there exists a number $\eta\in (y,x)$ such that 
$$\frac{\cos(x)-\cos(y)}{x^2-y^2}=-\frac{\sin(\eta)}{2\eta}$$
A: Suppose $x \neq y$, without lost of generality we could assume that $y<x$, then Cauchy's mean value theorem states that if $f$ and $g$ are continuous on the interval $[x,y]$, and $f$ and $g$ are differentiable on $(x,y)$ then there is $c \in (x,y)$ such that:
$$ \frac{f(y)-f(x)}{g(y)-g(x)}=\frac{f'(c)}{g'(c)} $$
In this case we have that $f(x)=cos(x)$ and $g(x)=x^2$, then there is $c \in (y,x)$:
$$\frac{cos(x)-cos(y)}{x^2-y^2}=\frac{-sin(c)}{2c} $$
Then if $(x,y) \to (0,0)$ since $c \in (x,y)$ then $c \to 0$:
$$lim_{(x,y) \to (0,0)} \frac{cos(x)-cos(y)}{x^2-y^2} = lim_{c \to 0}\frac{-sin(c)}{2c}=\frac{-1}{2} lim_{c \to 0}\frac{sin(c)}{c}=\frac{-1}{2}$$
Because $lim_{c \to 0}\frac{sin(c)}{c} = 1$
