Find if it exists the limit : $$\lim_{(x,y)\to(0,0)}\frac{x^4y^4}{(x^2+y^4)^3}$$

I've tried the following :

1st attempt : Using polar coordinates:

Set $x= r\cos\theta$ and $y=r\sin\theta$

$$\frac{x^4y^4}{(x^2+y^4)^3}=\frac{r^8\cos^4\theta \sin^4\theta}{(r^2\cos^2\theta+r^4\sin^4\theta)^3}=\frac{r^2\cos^4\theta\sin^4\theta}{(\cos^2\theta+r^2\sin^4\theta)^3}$$


Now the limit mentioned above would be equal to zero if and only if the denominator is different than zero.

Hence we need to calculate the limit in the case where $\theta = \frac{\pi}{2}+k\pi$ and compare it with the precalculated limit. However I was not able to get rid of the indeterminate form.

2nd attempt : Choosing a specific path $y = ax$.


$$\lim_{x\to0}\frac{a^4x^2}{(1+a^4x^2)^3}=0.$$No conclusion about the limit.

If anyone could give me hints or point me in the right direction I would be grateful.

Thanks in advance.


1 Answer 1


Approaching along the path $x = y^2$ (motivated by trying to simplify $x^2 + y^4$, we have

$$\lim_{y \to 0} \frac{(y^2)^4 y^4}{((y^2)^2 + y^4)^3} = \lim_{y \to 0} \frac{y^{12}}{(2y^4)^3} = \frac 1 8 \ne 0.$$

Hence, the limit does not exist.

Alternative interpretation in polar coordinates: $\cos \theta \to 0$ in such a way that the denominator vanishes at the same order that the numerator does, and so the limit is non-zero.

  • $\begingroup$ haven't you simply proven that the limit is 1/8? $\endgroup$
    – S. McGrew
    Commented Feb 28, 2018 at 19:24
  • $\begingroup$ @S.McGrew No, because there are obvious paths upon which the limit is $0$. $\endgroup$
    – user296602
    Commented Feb 28, 2018 at 19:28
  • $\begingroup$ So the trick to this is to get both the numerator and denominator down to the same degree so we could simplify (this is why you chose $x=y^2$ I guess).However I'm not sure how to get both numerator and denominator down to the same degree when we express them in polar coordinates, should i use another change of variables? $\endgroup$
    – Raku
    Commented Feb 28, 2018 at 19:32
  • $\begingroup$ @Raku Yes, that is why I chose $x = y^2$. This isn't easily terribly transparent in polar coordinates, so it's best to just leave it in the standard coordinates for showing that this limit doesn't exist. $\endgroup$
    – user296602
    Commented Feb 28, 2018 at 19:33
  • $\begingroup$ Note that from the AM-GM inequality, $x^2+y^4\ge 2\sqrt{x^2y^4}$ with equality when $x^2=y^4$, or $x=\pm y^2$. $\endgroup$
    – Mark Viola
    Commented Feb 28, 2018 at 19:53

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