How can I get the limit of this problem How to prove that:
  $$\lim_{(x,y) \rightarrow (0,0)}  (x+y)\sin(\frac{1}{x})\sin(\frac{1}{y})=0$$


I want to know how to do it. I'd appreciate it a lot.
So we have, $$ f(x,y)=(x+y)\sin(\frac{1}{x})\sin(\frac{1}{y})$$
then, using the fact that $\forall u\in\Bbb{R},|\sin(u)|&lt;1$, $$ 0&lt;|f(x,y)|=|(x+y)\sin(\frac{1}{x})\sin(\frac{1}{y})|\leq|x+y|$$
Thus,
$$ 0&lt;|f(x,y)|\leq |x+y|\xrightarrow[(x,y)\rightarrow(0,0)]{}0$$
By the sandwich theorem we have the answer requested.
\lim_{x,y\to 0,0}(x+y)\sin1/\xsin1/y=0
   \lim_{x\to 0}(x+y)[cos(x-y)/cosxcosy]=0
appling limit:-
  \lim_{x\to 0}(x+y)][\lim: \lim_{x\to 0}[\cos(x-y)/\cosxcosy]]=0
  [0+0][cos(0-0)/cos0cos0]=0
  0*[cos0/1*1]=0
  0*[1/1]=0
  0=0
  proved