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Consider the function $f : \mathbb{R}^2 \to \mathbb{R}$ defined by

$$ f(x,y) = \begin{cases} \frac{xy(x^2-y^2)}{x^2+y^2} & \text{for $(x,y) \neq (0,0),$}\\ 0 & \text{for $(x,y) = (0,0).$} \end{cases} $$

I am hoping to show this is continuous, but I do not know how to handle multivariate limits, can I simply take one and then the other or what do you do?

I have no $\varepsilon-\delta$ version of continuity at hand, only the usual limit formulation.

user280266
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    Have you tried polar coordinates? Then show it does not depend on theta as R tends to zero. – Jean-François Gagnon Oct 14 '15 at 15:39
  • @Jean-FrançoisGagnon Nope, but I will give it a shot. I did not find it looked like an expression that would be easily translated to polar coordinates. – user280266 Oct 14 '15 at 15:45
  • Well the denominator is r^2 so maybe... – Jean-François Gagnon Oct 14 '15 at 15:46
  • @Jean-FrançoisGagnon Yea, I know, but that just means that the angle has to cancel from the top where only things are multiplied together... – user280266 Oct 14 '15 at 15:47
  • if by "...take one and then the other" you mean first take limit as $x \rightarrow 0$ and take limit as $y \rightarrow 0$ then no you can't do that. See http://math.stackexchange.com/questions/1407614/lim-y-rightarrow-b-lim-x-rightarrow-a-f-neq-lim-x-y-rightarrow-a/1408783#1408783 – R_D Oct 14 '15 at 15:48
  • @user280266 Could you give us "the usual limit formulation"? – Luiz Cordeiro Oct 14 '15 at 16:17

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$$|f(x,y)|=\frac{|xy||x^2-y^2|}{x^2+y^2}\leq\frac{|xy|(x^2+y^2)}{x^2+y^2}=|xy|$$ which goes to zero as $(x,y)$ goes to zero.

Luiz Cordeiro
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