<s>[INST] Is $\displaystyle\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}$ equal to $-\dfrac 12$ or just $-\dfrac{\sin y}{2y}$ Question:
$$\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}=?$$
Here is my try:
\begin{align}\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}&=\lim_{x\rightarrow y}\dfrac{-2 \sin (\frac 12(x+y))\sin (\frac{1}{2}(x-y))}{(x+y)(x-y)}\\
&=-\dfrac{2 \sin y}{2y}\dfrac 12\\
&=-\dfrac{\sin y}{2y}\end{align}
My question: Is $-\dfrac{\sin y}{2y}$ the final answer or can it be calculated further as $-\dfrac12$?

I also try different route:
Let $p=x-y$ so $x=p+y$ and $p\rightarrow 0$ as $x \rightarrow y$. Hence,
\begin{align}\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}&=\lim_{p\rightarrow 0}\dfrac{\cos (p+y)-\cos y}{p^2+2py+y^2-y^2}\\\\
&=\lim_{p\rightarrow 0}\dfrac{-2 \sin (\frac{1}{2}(p+2y))\sin (\frac{1}{2}p)}{p(p+2y)}\\\\
&=\lim_{p\rightarrow 0}\dfrac{-2 \sin (\frac{1}{2}(p+2y))}{(p+2y)} \dfrac{\sin (\frac{1}{2}p)}{p}\\\\
&=-\dfrac{2\sin y}{2y} \dfrac{1}{2}\\\\
&=-\dfrac{\sin y}{2y}\end{align}
Okay, so that left me with the same result. What is the correct final answer, $-\dfrac{\sin y}{2y}$ or $-\dfrac 12$?
Thanks
 [/INST] we know by definition that 
$$\lim_{x\rightarrow y}\left(\frac{f(x) - f(y)}{x - y}\right) = \frac{df}{dy}$$
so the given can be expressed as
$$\lim_{x\rightarrow y}\left(\frac{\cos(x) - \cos(y)}{(x - y)(x + y)}\right)$$
that is equal to
$$\left(\frac{df}{dy}\right)\left(\lim_{x\rightarrow y}\frac{1}{x + y}\right)$$
at last becomes equal to
$$\frac{- sin(y)}{2y}$$
</s>