I've tried to get the following limit:

$$ \lim_{(x,y)\rightarrow(0,0)} xy\sin\left(\frac{1}{xy}\right) $$

wolfram claims it doesn't exist.

  1. How to show that?
  2. Why can't I take $z=xy$ and receive a known limit of one variable?

$$ \lim_{z\rightarrow0} z \sin\left(\frac{1}{z}\right) $$

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    $\begingroup$ Could it be because Wolfram is concerned that $\lim_{z\rightarrow0} zi \sin(\dfrac{1}{zi}) \neq \lim_{z\rightarrow0} z \sin(\frac{1}{z})$. That might explain the error Wolfram gives $\endgroup$ – user73445 May 4 '13 at 22:07
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    $\begingroup$ Is there a way to specify that $x,y$ are real when using wolfram? $\endgroup$ – TCL May 4 '13 at 22:26
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    $\begingroup$ It also depends on exactly how you define limits. (If you require the function to be defined in a full punctured neighborhood of the points you approach or not.) See math.stackexchange.com/questions/283747 for some more discussions. $\endgroup$ – mrf May 4 '13 at 22:33
  • $\begingroup$ @TCL Surprisingly it doesn't seem to allow the usual Mathematica syntax Assuming[Element[x | y, Reals], Limit[x*y*Sin[1/(x*y)], x->0, y->0]]. Although when I run the query even without Assuming it gives me 0 as the result with a footnote assuming x,y are reals. $\endgroup$ – Voo May 14 '14 at 11:47

1) The limit clearly exists since $\sin$ is bounded and $xy\rightarrow 0$.

2) In this problem you can. In a problem such as $\lim_{(x,y)\rightarrow (0,0)}\frac{x}{y}$ you run into trouble because you seem to suggest that it's ok to write it as $\lim_{z\rightarrow 0}\frac{z}{z}=1$ and that's wrong. To see why it's wrong, $x/y$ can be arbitrarily large if you keep $x$ fixed and make $y$ smaller.

  • $\begingroup$ How to know when it's ok to take such $z$? or polar coordinates? $\endgroup$ – User May 4 '13 at 22:27

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