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I have $f(x,y) =\frac{2xy^2 }{x^2+y^4}$ if $(x,y) \neq (0,0)$ and $0$ if $(x,y) = (0,0)$. I figured I could just find that the limit as $\frac{2xy^2 }{x^2+y^4}$ does not equal to $0$, which should show discontinuity, but it does and I don't know how else one can show discontinuity. I showed it was not differentiable with the definition of the derivative if that helps at all, but I can't reuse that. Thanks for any help.

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"but it does": But what does what? If the limit of $f(x,y)$ as $(x,y)\to (0,0)$ were to exist and be $0$, then the function would be continuous. But the limit does not exist. – Jonas Meyer Sep 26 '12 at 4:59

By taking the path $\mathbf{r}_a(t)=(t,\sqrt{\frac{t}{a}})$ reaching to $(0,0)$, you will find out that the limit is not exist at the origin. In fact, it depends on the value of $a$.

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