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Show that the Limit doesn't exist

$\lim \limits_{(x,y,z) \to (0,0,0)} \frac{xyz}{x+y+z}$

I try to replace all of $x$, $y$, and $z$ with what they are actually approaching. But I'm stuck there and I can't prove that the Limit doesn't exist

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  • $\begingroup$ Try to evaluate the function along the line $x=y=-z/2$. $\endgroup$ – Oscar Lanzi Sep 17 at 1:29
  • $\begingroup$ The function isn't defined when $x+y+z=0$. As you approach $(0,0,0)$ near that plane, you can make the function do crazy stuff. $\endgroup$ – yoyo Sep 17 at 1:34
  • $\begingroup$ @OscarLanzi That line is not in the domain. But it's not hard to see that in $D=x+y+z\ne 0$ that this function is unbounded in every $D\cap B(0,r).$ $\endgroup$ – zhw. Sep 17 at 1:50
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An approach to prove that a limit does not exist, is to choose various lines / curves that run through the point under consideration. If the limit exist it exists on all paths through that point. If you can find 2 that give different results, you are done.

We have a problem along the path $(t,t,-2t),$ and all curves crossing that line.

This will not rigorously prove that a limit does exist, though.

Another way to do these sorts of problems that is pretty easy to consistently apply is to convert to polar / spherical coordinates.

Then you get $\frac {r^m f(\theta,\phi)}{r^n g(\theta,\phi)}$ Then if there exists some $\theta,\phi$ combination, such that the denominator is zero (and the numerator non-zero), you have a problem, and the limit does not exist. If that is not a problem, then you need to show that the degree of $r$ in the numerator is greater than the degree of $r$ in the denominator.

And finally, you have a lot of latitude to chose a distance metric that is easy to work with. You are not obligated to use the Euclidean metric.

$d((x,y,z),(0,0,0)) = |x|+|y|+|z|$ or $\max(|x|,|y|,|z|)$ might be easier to work with.

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How does the function behave on the path $(t+t^4,t,-2t)$ as $t\to 0^+?$

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