2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ HINT: $|2\sin(x-y)|=\left|\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)\right|\le \left|\sin\left(\frac{x-y}{2}\right)\right|$ $\endgroup$ – d80d2729a352b1366139fc119d3345 Aug 30 '14 at 11:50
  • $\begingroup$ Hint: $$\sin(x)-\sin(y)=\cos(\theta x+(1-\theta)y)(x-y)$$ $\endgroup$ – pointer Aug 30 '14 at 12:06
3
$\begingroup$

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 $\cos 1$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Ali Caglayan Aug 30 '14 at 12:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.