# 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?

• HINT: $|2\sin(x-y)|=\left|\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)\right|\le \left|\sin\left(\frac{x-y}{2}\right)\right|$ – d80d2729a352b1366139fc119d3345 Aug 30 '14 at 11:50
• Hint: $$\sin(x)-\sin(y)=\cos(\theta x+(1-\theta)y)(x-y)$$ – pointer Aug 30 '14 at 12:06

## 1 Answer

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 $\cos 1$.

• Strictly speaking the second last line should have the 1 on the outside. This may cause confusion as you cannot evaluate mini limits inside a limit. – Ali Caglayan Aug 30 '14 at 12:11