Calculating the value of the limit $$\lim_{(x,y)\to(1,2)}(\sin(y)-\sin(x))$$
My try:
I got as $$\sin(2)-\sin(1)$$
But I cannot calculate the exact value of the given limit. Can anyone please explain this.
 A: You did the work properly.
Now, if you want a number, remember that $\frac \pi 3 \approx 1$ which would make $0$ as an approximation.
Now, if you want "better", use the Taylor series
$$\sin(x)=\frac{\sqrt{3}}{2}+\frac{1}{2} \left(x-\frac{\pi
   }{3}\right)+O\left(\left(x-\frac{\pi }{3}\right)^2\right)$$ which makes
$$\sin(1)\approx \frac{1}{2}+\frac{\sqrt{3}}{2}-\frac{\pi }{6}$$
Similarly
$$\sin(x)=\frac{\sqrt{3}}{2}-\frac{1}{2} \left(x-\frac{2 \pi
   }{3}\right)+O\left(\left(x-\frac{2 \pi }{3}\right)^2\right)$$ which makes
$$\sin(2)\approx -1+\frac{\sqrt{3}}{2}+\frac{\pi }{3}$$ Then
$$\sin(2)-\sin(1)\approx \frac {\pi-3}2=0.0708$$ while the exact value would be $0.0678$.
A: You have done the limit correctly. If this is for school, your teacher surely doesn't expect you to know trigonometric values for odd degrees, so you're fine. Unless it's for multiples of 15 degrees (that is, multiples of pi/6 radians), you probably aren't ever going to be expected to calculate them offhand.
A: Yes your result is correct, indeed since the function is continuous at the point we have
$$\lim_{(x,y)\to(1,2)}(f(x,y))=f(1,2)$$
that is 
$$\lim_{(x,y)\to(1,2)}(\sin(y)-\sin(x))=\sin 2 -\sin 1$$
where $2$ and $1$ are expressed in radians, a numerical evaluation leads to $\approx 0.068$.
