Evaluate $lim_{(x,y)\to(0,0)\frac{xy}{y-x^3}}$

1. Oct 10, 2015

duarthiago

1. The problem statement, all variables and given/known data
Evaluate $lim_{(x,y)\to(0,0)\frac{xy}{y-x^3}}$

2. Relevant equations

3. The attempt at a solution
I've tried move towards the origin along the path $\gamma_{1} (t) = (t,2t^3)$, which leads to $2t=0$. Then I've tried to use $\gamma_{2} (t) = (0,t)$ which leads to $\frac{0}{t}$, then applying the L'Hospital's rule this limit is 0 too (can I always do it when I have a single-variable limit?).

I've suspected that the limit exists and its value is 0, then I've tried to use polar coordinates:

$lim_{(x,y)\to(0,0)\frac{xy}{y-x^3}}$

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

But I've got stuck there because I can't figure how to show that $\frac{cos(k)sin(k)}{sin(k)-r^2 cos^3 (k)}$ is bounded.

A further question: It is related to the question on L'Hospital's rule, above. In regards to the limit

$lim_{(x,y) \to (0,0)} \frac{xy(x-y)}{x^4 + y^4}$
The path $\gamma_{A} (t) = (t,0)$ leads to 0.
$\gamma_{B} (t) = (t,2t)$ leads to $\frac{-2t^3}{17t^4}=\frac{-2}{17t}$. Can I tend t to zero, conclude that the limit goes to the minus infinity and because of that say that the original limit doesn't exist? Maybe it seems a silly question, but until now I haven't seen something like this happening. It is always something bounded multiplied by zero.

2. Oct 10, 2015

duarthiago

Well, I worked a little bit more on the problem and polar coordinates wasn't necessary. Taking a curve like $\gamma (t) = (\sqrt[3]{t-t^2},t)$ is enough.
About my second question: now is quite clear that if I choose a curve that goes through the origin, then I have to study what happens when t -> 0 to conclude something about the behavior of the original function when (x,y) -> (0,0).

3. Oct 10, 2015

pasmith

You don't need to resort to L'Hopital's theorem when the numerator is constantly zero: the limit will be zero.

The obvious problem here - which you do not mention - is that $f(x,y) = \frac{xy}{y - x^3}$ is not defined on the curve $y = x^3$, which of course passes through the origin.

It isn't: For each $r$, it is unbounded when $\sin k = r^2 \cos^3 k$. But what you actually want is the limit of $r$ times that expression. That might be bounded as $r \to 0$, even if $\frac{cos(k)sin(k)}{sin(k)-r^2 cos^3 (k)}$ is not. But even that would not be sufficient to conclude that the limit of $f$ exists: it only considers what happens as you approach along straight lines, and there are other ways of approaching the origin.

Exercise: Set $\delta > 0$ and consider the line $y = \delta - x$ for $0 < x < \delta$. This lies inside the open ball $\|(x,y)\| < \delta$. Look at $f(x,\delta - x)$. What happens to the sign of the denominator between $x = 0$ and $x = \delta$?

What can you then conclude about the existence of the limit of $f$ at the origin?

Yes.

4. Oct 10, 2015

duarthiago

Thank you for your insightful answer. About the exercise, I'm not sure, but it seems that $y$ goes through (0,$\delta$) and ($\delta$,0). Then $x^3$ (which also lies on the circle) will be greater than $\delta - x$ at some point, so the sign of the denominator becomes negative. $y$ won't pass at the origin, but the sign will always change no matter how small delta is. So, can I conclude the limit of $f$ at origin doesn't exist precisely because of that behavior?