How do I prove that $f(x,y)=xy$ is not uniformly continuous? I can prove that it is continuous, because as long as $$\left|\sqrt{(x-x')^2+(y-y')^2}\right|<\min{\left\{\frac{\epsilon}{2(1+|x|+|y|)},1\right\}}$$, we have $|xy-x'y'|<\epsilon$.
But how do I prove that it is not uniformly continuous?
 A: HINT: Notice that the $\delta$ that you used to prove continuity depends heavily on $x$ and $y$: when either of them is large in absolute value, your $\delta$ is very small compared with $\epsilon$. Try to make that observation more rigorous: show that that there is an $\epsilon>0$ such that for any $\delta>0$, if you take $|xy|$ large enough, you can find a point $\langle x',y'\rangle$ such that $\sqrt{(x-x')^2+(y-y')^2}<\delta$, but $|xy-x'y'|\ge\epsilon$. That will show that there is no $\delta$ that ‘works’ for that $\epsilon$ independently of the choice of point $\langle x,y\rangle$. (Don’t work too hard to pick this ‘bad’ $\epsilon$; most anything will work!)
A: You want to show that there is some $\epsilon>0$ for which, for any $\delta>0$, there is some $x,y,x',y'$ with
$$\sqrt{(x-x')^2+(y-y')^2}<\delta \;\text{ yet }\; |xy-xy'|\geq \epsilon.$$ Let $\epsilon=1/2$ and note that
$|xy-xy'|\geq |x||y-y'|$ so for any $\delta>0$, if $|x|>1/\delta$ we have 
$$|y-(y+\delta/2)|=\delta/2<\delta\; \text{ yet }\;|xy-x(y+\delta/2)|=|x|\delta/2>1/2$$
hence $xy$ is not uniformly continuous.
