Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

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

share|cite|improve this question
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 – user73445 May 4 '13 at 22:07
Is there a way to specify that $x,y$ are real when using wolfram? – TCL May 4 '13 at 22:26
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 for some more discussions. – mrf May 4 '13 at 22:33
@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. – Voo May 14 '14 at 11:47
up vote 1 down vote accepted

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.

share|cite|improve this answer
How to know when it's ok to take such $z$? or polar coordinates? – User May 4 '13 at 22:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.