<s>[INST] Evaluating $\lim\limits_{(x,y)\rightarrow(1,1)} \frac {\sin(x) - \sin (y)} {x-y}$ I am taking a calculus exam in less than one week, and I've stumbled upon this expression.

$$\lim\limits_{(x,y)\rightarrow(1,1)} \frac {\sin(x) - \sin (y)} {x-y}$$

Which I know to be cos(1), but I cannot seem to find the inequalities to make an $\epsilon-\delta$ proof of said limit.
I've tried coordinate change and Taylor, to no avail. Whenever I do Taylor(1) or equivalent infinitesimals, any variable I have manages to cancel out. If I don't, the thing just grows and grows...
Is there something I am entirely missing from the start?
 [/INST] You can use this identity:
$$\sin x - \sin y = 2\sin \frac{x - y}{2}\cos \frac{x + y}{2}.$$
Then your limit becomes:
$$\lim_{(x, y) \to (1,1)} \frac{2\sin \frac{x - y}{2}\cos \frac{x + y}{2}}{x - y} =$$
$$2 \lim_{(x, y) \to (1,1)} \frac{\sin \frac{x - y}{2}\cos \frac{x + y}{2}}{x - y} =$$
$$\frac{1}{2} \cdot2 \lim_{(x, y) \to (1,1)} \frac{\sin \frac{x - y}{2}\cos \frac{x + y}{2}}{\frac{x - y}{2}} =$$
$$\lim_{(x, y) \to (1,1)} \frac{\sin \frac{x - y}{2}}{\frac{x - y}{2}} \cdot \cos \frac{x + y}{2}=$$
$$\lim_{(x, y) \to (1,1)} 1 \cdot \cos \frac{x + y}{2}=$$
$$\lim_{(x, y) \to (1,1)} \cos \frac{2}{2},$$
which exactly yields $\cos 1$.
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