Does $ \lim_\limits {(x,y)\to (0,0 )} \frac{\sin (x) -\sin (y)}{x + y}$ exist? Does $$\lim_{(x,y)\to (0,0 )} \frac{\sin (x) - \sin (y)}{x + y}$$ exist?
It's clear that the $\sin x$ will approach $0$ as $x$ or $y$ is approaching 0. Should I maybe use to find the answer polar coordinates? Where $x = r \cos \varphi$,$y = r \sin \varphi$ and so $r^2=x^2+y^2$.
I would appreciate any kind of help.
 A: What happens when you approach the origin along the curve $(t, t^2)?$ What about $(t^2, t)?$
A: No. Along x=y it is zero and at x=-y it is not defined.
Take $x=-y+y^2$. Then, as $y \to 0$,
 it is about $2y/y^2 = 2/y \to \infty$.
A: Try approaching (0,0) along the x-axis, then try again approaching along the y-axis.
Ced
A: Hint
As pointed out in the comments:
$$\frac{\sin(t)-\sin(t)}{2t} \to 0$$
$$\frac{\sin(t)-\sin(-t)}{2t}=\frac{\sin(t)+\sin(t)}{2t} \to 1/2$$
A: We know from the definition of the $\mathtt{sinc}$-function (unnormalized version) $\mathtt{sinc}\left(x\right)\;:=\;\frac{\sin\left(x\right)}{x}$ that
$$\mathtt{lim}_{x\to 0}\;\mathtt{sinc}\left(x\right)=\frac{\sin\left(x\right)}{x}=1$$
Now lets call your problem function $f$:
$$f\left(x,y\right)\;:=\;\frac{\sin\left(x\right)-\sin\left(y\right)}{x+y}$$
We have for $y=0$:
$$\mathtt{lim}_{x\to 0}\;f\left(x,0\right)=\mathtt{lim}_{x\to 0}\;\frac{\sin\left(x\right)}{x}=1$$
But for $x=0$:
$$\mathtt{lim}_{y\to 0}\;f\left(0,y\right)=-\;\mathtt{lim}_{y\to 0}\;\frac{\sin\left(y\right)}{y}=-1$$
So a limit does not exist.
