# Showing $\lim_{(x,y)\to (0,0)}\frac{xy^4}{x^2+y^8}$ Exists and Does Not Exist…

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?

• Yes, dont trust mathematica or wolframalpha. – Masacroso Feb 24 '17 at 4:29
• 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. – lordoftheshadows Feb 24 '17 at 4:30
• @Masacroso What about polar? – The Substitute Feb 24 '17 at 4:30
• @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 $r$. – lordoftheshadows Feb 24 '17 at 4:36
• – Robert Howard Jul 20 '18 at 17:32