Continuous function and limit $ f(x,y)=\left\{\begin{matrix} (x-y)\sin\frac{1}{x}\sin\frac{1}{y} & , xy\neq 0 \\ 0 &, x=y=0 \end{matrix}\right. $ I have this function:$$ f(x,y)=\left\{\begin{matrix}
(x-y)\sin\frac{1}{x}\sin\frac{1}{y} & ,  xy\neq 0 \\ 0
 &, x=y=0
\end{matrix}\right. $$
a) Show that $ \lim_{x\rightarrow 0}[\lim_{y\rightarrow 0} f(x,y)] $ does not exist.
b) Show that $ f(x,y) $ is continuous at $ (0,0) $ .
For (a) I took $ \lim_{y\rightarrow 0} f(x,y) $ and I got that it's equal to $$ x\sin\frac{1}{x}\lim_{y\rightarrow 0}\sin\frac{1}{y} - \sin\frac{1}{x}\lim_{y\rightarrow 0}y \sin\frac{1}{y} $$, but $$ \lim_{y\rightarrow 0}\sin\frac{1}{y} $$ does not exist, so $ \lim_{y\rightarrow 0} f(x,y) $ does not exist and $ \lim_{x\rightarrow 0}[\lim_{y\rightarrow 0} f(x,y)] $ does not exist. Am I right?
For (b) I guess that I have to show that $$ \lim_{(x,y)\rightarrow (0,0)} f(x,y) = 0 $$ How can I do this ?

$sin{(1/x)sin(1/y)}$ is an interval and $x-y$ under the limit of $(x,y)\to{(0,0)}$ is an interval too. if a function in a close interval is continuous, its valued field is an interval, but its converse proposition may be not true. for example, your function is a counterexample. so you should separate your problem into two cases with $f^{'}(0,0)$ and $lim_{(x,y)\to{(0,0)}}f^{'}(x,y)$
the first limit exists but the second do not! for example rewrite your function to $F_{x}^{'}(x,y)=sin(1/x)sin(1/y)+(x-y)sin(1/y)cos(1/x)(-1/x^2)$
$F_{y}^{'}(x,y)=-sin(1/x)sin(1/y)+(x-y)sin(1/x)cos(1/y)(-1/y^2)$
divide both side with $sin(1/x)sin(1/y)$ to get
$1+(x-y)cos(1/x)(-\frac{1}{x^{2}sin(1/x)}）$
$-1+(x-y)cos(1/y)(-\frac{1}{y^{2}sin(1/y)）}$
since $f(x)=x^2{sin(1/x)}$ or equal to zero exists a limit at $f^{'}(0)$, but $lim_{x\to{0}}f^{'}(x)=2xsin(1/x)-cos(1/x)$ do not exists, this result give a certain conclusion to your problem. thank you!
