2

Consider the limit $$\lim_{(x,y)\to (0,0)}\frac{xy^4}{x^2+y^8}$$

Although Wolfram Alpha and Mathematica say that the limit is zero, consider the limits along the paths $y=0$ and $x=y^4$.

If $y=0$, then the limit becomes $$\lim_{x\to 0}\frac{0}{x^2}=0$$

However, if $x=y^4$, then the limit becomes $$\lim_{y\to 0}\frac{y^8}{2y^8}=1/2$$

Therefore, the limit does not exist.

On the other hand, when I switch to polar coordinates, $x=r\cos\theta, y=r\sin\theta$, the limit becomes $$\lim_{r\to 0^+}\frac{r^5\cos\theta \sin^4\theta}{r^2\cos^2\theta +r^8\sin^8\theta }=\lim_{r\to 0^+}\frac{r^3\cos\theta\sin^4\theta}{\cos^2\theta+r^6\sin^8\theta}=\frac{0}{\cos^2 \theta}$$

Case 1: If $\cos^6 \theta \not=0$, then the limit is zero.

Case 2: If $\cos^6\theta=0$, then (since this implies $\sin^8\theta=1$) the limit is $$\lim_{r\to 0}\frac{r^3 (0) (1)}{0+r^6 (1)}=\lim_{r\to 0}\frac{0}{r^6}=0$$

Therefore, according to my work in polar coordinates and Mathematica the limit is zero. However, I have found two different paths $y=0$ and $x=y^4$ for which the limits are distinct.

Does this limit exist, and where is the error?

Matthew Cassell
  • 4,364
  • 4
  • 23
  • 31
The Substitute
  • 4,098
  • 1
    Yes, dont trust mathematica or wolframalpha. –  Feb 24 '17 at 04:29
  • 1
    This limit pretty obviously doesn't exist. Consider approaching from $y=x$ then the limit diverges and you're done. Lots of computer algebra systems are terrible at finding limits of multivariable functions precisely because you can screw up very easily. – Zaros Feb 24 '17 at 04:30
  • @Masacroso What about polar? – The Substitute Feb 24 '17 at 04:30
  • Can you please elaborate on the $\cos^6(\theta) \neq 0$ parts.. it doesn't make much sense to me. – Ahmed S. Attaalla Feb 24 '17 at 04:31
  • @AhmedS.Attaalla If $\cos^6$ is nonzero, then the numberator is zero while the denonomator is nonzero, so the entire fraction is zero. In case 2, I argued that the limit is still 0 even when $\cos^6 = 0$. – The Substitute Feb 24 '17 at 04:32
  • @TheSubstitute the paths that you chosen from polar coordinates converge to zero but you previously shown that the limit can be $1/2$, this is enough to see that the limit doesnt exists. –  Feb 24 '17 at 04:33
  • @Masacroso I don't want the final answer to depend on the technique I am using. – The Substitute Feb 24 '17 at 04:33
  • 6
    @TheSubstitute the reason your work with polar coordinate doesn't work is you're assuming that $\theta$ is constant which is not correct. $\theta$ is allowed to vary as well as $r$. – Zaros Feb 24 '17 at 04:36
  • @TheSubstitute a general technique to prove that these limits doesnt exists is define a variable $z:=y/x$, then $z\to c$ for any $c$ at the same time that $(x,y)\to (0,0)$, then you can play with this kind of sequences. See here. But setting $x=y^4$ is enough as you shown before and simple. –  Feb 24 '17 at 04:36
  • @TheSubstitute This is one of those rare examples where the limit along any straight line into the origin is the same and the limit still does not exist. Your methods do not conflict, the second is just not strong enough to prove that the limit exists (it's great for proving it doesn't exist when it works, though). – spaceisdarkgreen Feb 24 '17 at 04:40
  • 2
    Possible duplicate of Limit $\frac{x^2y}{x^4+y^2}$ is found using polar coordinates but it is not supposed to exist. – Robert Howard Jul 20 '18 at 17:32

0 Answers0