MHB Does the Multivariable Limit of xy^8/(x^3+y^12) Exist?

[Prove It]

Hi everyone. A friend of mine asked for help evaluating this multivariable limit.

$\displaystyle \begin{align*} \lim_{(x,y) \to (0,0)} \frac{x\,y^8}{x^3 + y^{12}} \end{align*}$

We got the answer of 0 by converting to polars.

$\displaystyle \begin{align*} \lim_{(x,y) \to (0,0)} \frac{x\,y^8}{x^3 + y^{12}} &= \lim_{r \to 0} \frac{r\cos{( \theta )} \, \left[ r\sin{ (\theta ) } \right] ^8}{ \left[ r\cos{ (\theta ) } \right] ^3 + \left[ r\sin{ ( \theta ) } \right] ^{12} } \\ &= \lim_{r \to 0} \frac{r^9 \cos{ ( \theta ) } \sin^8{ ( \theta ) } }{r^3 \cos^3{( \theta ) } + r^{12} \sin^{12}{ ( \theta ) } } \\ &= \lim_{r \to 0} \frac{r^6 \cos{( \theta ) } \sin^8{ ( \theta ) } }{ \cos^3{( \theta )} + r^9 \sin^{12}{(\theta ) } } \\ &= \frac{0}{\cos^3{(\theta ) } } \end{align*}$

Now for all $\displaystyle \begin{align*} \theta \neq \frac{ \left( 2n + 1 \right) \, \pi}{2}, n \in \mathbf{Z} \end{align*}$, we have $\displaystyle \begin{align*} \frac{0}{\cos^3{( \theta ) }} = 0 \end{align*}$ and since $\displaystyle \begin{align*} \theta = \frac{ \left( 2n + 1 \right) \, \pi}{2} \end{align*}$ are all REMOVABLE discontinuities (i.e. holes), that means $\displaystyle \begin{align*} \lim_{\theta \to \frac{ \left( 2n + 1 \right) \, \pi}{2}} \frac{0}{\cos^3{( \theta ) }} = 0 \end{align*}$ as well, thereby still approaching the same value from those paths too...

Therefore surely $\displaystyle \begin{align*} \lim_{(x,y) \to (0,0) } \frac{x\,y^8}{x^3 + y^{12}} = 0 \end{align*}$.

Wolfram appears to agree with us too.My friend's online assessment however has the answer of DNE. When asked about it and showing our work and Wolfram's output, my friend's lecturer is still adamant that the limit does not exist, because the value we get when converting to polars and letting r approach 0 isn't always defined i.e. where $\displaystyle \begin{align*} \theta = \frac{ \left( 2n + 1 \right)\, \pi}{2} \end{align*}$. But wouldn't that just mean that there's more paths that need to be tested, and as we have shown, we still approach the same value as $\displaystyle \begin{align*} r \to 0 \end{align*}$ from all paths with $\displaystyle \begin{align*} \theta \neq \frac{ \left( 2n + 1 \right) \, \pi}{2} \end{align*}$ as we do when $\displaystyle \begin{align*} \theta = \frac{ \left( 2n + 1 \right) \, \pi}{2} \end{align*}$.Would anyone here be able to go over our work and verify who is correct in this case please?

 

[Fantini]

Your friend's lecturer is correct. Consider the path $x=y^4$. Then

$$\frac{xy^8}{x^3+y^{12}} = \frac{y^4 y^8}{(y^4)^3 + y^{12}} = \frac{y^{12}}{2y^{12}} = \frac{1}{2},$$

and the limit is $1/2$.

Best wishes. :)

 

[Prove It]

Your friend's lecturer is correct. Consider the path $x=y^4$. Then

$$\frac{xy^8}{x^3+y^{12}} = \frac{y^4 y^8}{(y^4)^3 + y^{12}} = \frac{y^{12}}{2y^{12}} = \frac{1}{2},$$

and the limit is $1/2$.

Best wishes. :)

A beautiful counterexample, how did you come up with it? :)

 

[Euge]

Hi Prove It,

Since Fantini has already displayed a counterexample, I'm giving a small remark that I hope will help. Looking at your work, I see that after changing to polar coordinates, you assumed $\theta$ is constant and found the limit to be zero. However, finding a bivariate limit by changing to polar coordinates does not always work, since it's possible to have the limit exist along every ray through the origin, but fail to exist in the general sense.

Here's a rigorous statement regarding the polar method. Let $f : \Bbb R^2 \to \Bbb R$. Suppose there is an $L$ such that to every $\epsilon > 0$, there corresponds a $\delta > 0$ such that for all $r$ and $\theta$, $|r| < \delta$ implies $|f(r,\theta) - L| < \epsilon$. Then

$$\lim_{(x,y)\to (0,0)} f(x,y) = \lim_{r\to 0} f(r,\theta) = L.$$

 

[Fantini]

This is a variation of a known limit that seemingly exists for all lines, such as when you tested polar coordinates, but when tested along other curves (like parabolas or the sort) does not exist. :) The common form is

$$\frac{xy^2}{x^2+y^4}.$$

Thank you for the compliment. Best wishes. :)