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Suppose a function $f(x, y)$ is defined as follows like this: $f(x, y)=\frac{xy^3}{x^3+y^4}$ when $(x, y)\neq (0, 0)$ and $f(x, y)=(0, 0)$ when $(x, y)=(0, 0)$. Is this function continuous at $(0, 0)$, please? I think it is. I have tried to let $y=kx$ and $y=x^a$. How can I prove it in general, please? I think I should show that $\lim f(x, y) = 0$. But how to do it in general, please?

user106035
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  • Show that the limit from the left and the limit from the right approach the same value. – David Nov 07 '13 at 03:43
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    How is the function defined for $(x,y)$ such that $x^3+y^4 = 0$? – copper.hat Nov 07 '13 at 03:49
  • This is not a one variable function. Hence the way you described does not work. – user106035 Nov 07 '13 at 03:50
  • @David The notion "left/right" for a function (and limit) of two variables is either ill-defined or not useful here. – zibadawa timmy Nov 07 '13 at 03:50
  • The question does not specify that point. – user106035 Nov 07 '13 at 03:51
  • Ah, yes, didn't see that. @zibadawatimmy, but we may generalize and state you one must show that from any direction the limit is the same then. – David Nov 07 '13 at 03:52
  • Copper has a good point. The function is not defined in a neighborhood of the origin – zibadawa timmy Nov 07 '13 at 03:53
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    @David I'm quite certain that is insufficient, and that there are two-variable rational functions whose limits exist in every direction and are the same but are nevertheless not continuous. Though lacking a calculus book, I cannot actually go looking for it, and I do not recall the function. So, grain of salt, maybe, but I'm pretty sure just coming from every direction isn't good enough. – zibadawa timmy Nov 07 '13 at 03:55
  • Then this should imply that this function is not continuous at the origin. Right? – user106035 Nov 07 '13 at 03:55
  • @user106035 Yes, it would. The creator of the problem may not have realized this problem. It is a very common issue when trying to create such a rational function with a not-so-obvious continuity question. – zibadawa timmy Nov 07 '13 at 03:56

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There are most likely simpler counterexamples, but this is easy to compute.

Even if we ignore that $f$ is not completely defined in a neighbourhood of the origin, the function $f$ is still not continuous.

$f(-\sqrt[3]{t^4},t+t^2) = \frac{-\sqrt[3]{t^4}(t+t^2)^3}{-t^4+(t+t^2)^4} = \frac{-\sqrt[3]{t^4} t^3 (1+t)^3}{t^5(t+2)(t^2+2t+2)} = -\frac{\sqrt[3]{t^4}}{t^5}\frac{(1+t)^3}{(t+2)(t^2+2t+2)}$.

We see that $\lim_{t \downarrow 0} f(-\sqrt[3]{t^4},t+t^2) = - \infty$, hence $f$ is not continuous at $(0,0)$.

copper.hat
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  • Indeed, if you limit along the lines $x=c\cdot y$ and $y=d\cdot x$, you will see that the limit does exist and is $0$ for these cases. Which means I remembered correctly, and the limit existing and being the same along every ray does not imply continuity. The notion of "direction" has to include the non-linear ones, like the parametrized curve that copper has given us here. – zibadawa timmy Nov 07 '13 at 04:08
  • Continuity in 2 dimensions means that for every $\epsilon$ there is a $\delta$ such that |f(x,y) - f(0,0)| < $\epsilon$ throughout a disk of radius $\delta$. This is a much stronger requirement than just coming in along some (or every) line, which only requires the inequality to hold on an interval. – Betty Mock Nov 07 '13 at 06:24