This question already has an answer here:

Here, if we go along $x=-y$ then the function itself is not defined, and limit seems to exist for all other paths of approach

Putting in another way, if we use polar coords, limit is:

$$\lim_{r\to 0} \frac{r^4 \sin^2\theta\cos^2\theta}{r^3(\sin^3\theta + \cos^3\theta)}$$

which is zero only when $(\sin^3\theta + \cos^3\theta)$ is nonzero, and function itself is not defined for values where $(\sin^3\theta + \cos^3\theta) = 0$

So here, what do we say? Does the limit exist? or does it not exist

(Additional Question)

Suppose that limit is being calculated $(x,y) \to (0,0)$. Then can we choose one of the paths as, say $x=0$, although in limit we only let variables to "tend" to a value? Is it allowed to use this path?


marked as duplicate by Nosrati, Lord Shark the Unknown, Brevan Ellefsen, user91500, Namaste Sep 4 '18 at 10:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ Yes, you can set $x=0$ and take a limit as $y\to 0$ $\endgroup$ – Jakobian Aug 30 '18 at 13:50
  • $\begingroup$ @Rumpelstiltskin Thanks a lot for necessary information :) $\endgroup$ – jeea Aug 30 '18 at 13:52

We have that

  • $x=y=t\to 0 \implies \frac{x^2y^2}{x^3+y^3}=\frac{t^4}{2t^3}=\frac t{2}\to 0$

  • $x=t\to 0 \quad y=-t+t^2\to 0 \implies \frac{x^2y^2}{x^3+y^3}=\frac{t^4-2t^5+t^6}{3t^4-3t^5+t^6}=\frac{1-2t+t^2}{3-3t+t^2}\to \frac13$

therefore the limit doesn't exist.

  • $\begingroup$ Thank you for good answer, my follow up question is whether checking along y=-x path allowed when function itself doesnt exist along that path? $\endgroup$ – jeea Aug 30 '18 at 15:00
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    $\begingroup$ @jeea No we can't assume $x=-y$ since thise points are excluded by the domain. You can note that my second path $x=t$ and $y=-t+t^2$ has been chosen to approach the line $x=-y$ but the quadratic term in $y$ has been added to avoid that the line is touched. $\endgroup$ – gimusi Aug 30 '18 at 15:03
  • $\begingroup$ @jeea When the limit doesn't exist, often this method allows to find easily those path with different limit with respect to the trivial path away from the non existence points (i.e. x=y in that case). $\endgroup$ – gimusi Aug 30 '18 at 15:06

The limit does not exist. Consider $x\to 0$ and $y$ such that $y^3=x^9-x^3$. Then $y\to 0$ but you can see using binomial series that $$\frac{x^2 y^2}{x^3+y^3}=\frac{x^2|x^9-x^3|^{2/3}}{x^9}\approx\frac{1}{x^5}(1-\frac23 x^6) $$ tends to infinity. Choosing $y^3=-x^9-x^3$ gives you a way to approach zero where the function tends to $-\infty$. So it does not have a limit.

For your additional question, if the limit exists you may calculate it along any path of your choice. However, you can't show that the limit exists by only considering certain choices of paths.

  • 1
    $\begingroup$ Note that $y^2 = (x^9-x^3)^{2/3}$. Your argument remains valid, of course. $\endgroup$ – Hans Engler Aug 30 '18 at 14:21
  • $\begingroup$ Fixed, thanks! I tried to have nice numbers, but failed :( $\endgroup$ – Kusma Aug 30 '18 at 14:31
  • $\begingroup$ Thanks for good answer, clears it up, follow up question is whether checking along x = -y allowed when function isnt even defined on that path? $\endgroup$ – jeea Aug 30 '18 at 15:01
  • $\begingroup$ You can't test where the function isn't defined, but it is usually interesting to look close to that set. $\endgroup$ – Kusma Aug 30 '18 at 15:12
  • $\begingroup$ @Kusma I agree, this answer by another user is confusing me with this subject, what I think his answer says is the opposite, I may be wrongly interpreting it: link $\endgroup$ – jeea Aug 30 '18 at 18:25

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