Nonexistence of the limit $\lim_{(x,y)\rightarrow (0,0)} \frac{x^2y^2}{x^3+y^3}$ How can we prove that the limit $\lim_{(x,y)\rightarrow (0,0)} \dfrac{x^2y^2}{x^3+y^3}$ doesn't exist?
I have tried a lot of different paths and all of them lead to zero. I have only tried paths belonging to the domain and so I thank you for the lights that you've thrown here for me. When I've plotted the graphic of the function, it was clear that this limit doesn't exist, but could not prove it. For instance, this is an exercise of the book of Louis Leithold.
 A: A good way is to find a path through the plane $(x,y)$ such that the limit is different from the limit from another path.
For exemple, following the path $y=x$ we have the limit of $0$. If you find a path with a different limit, then the limit doesn't exist.
A: I am going to essentially repeat nbubsis's answer, but with a little more explanation.  A lot of people (including me and probably nbubis) would say that the limit does not exist because there does not exist $r>0$ such that the function $x^2y^2/(x^3+y^3)$ is well-defined in a "deleted neighborhood" of $(0,0)$ of the form $\{(x,y) \mid 0< \sqrt{x^2+y^2}<r\}$ (nbubis explains why).  Other people would say you only have to consider pairs $(x,y)$ that belong to the domain of the function $f(x,y)=x^2y^2/(x^3+y^3)$.  This seems to be what Victor Chaves and kigen  are suggesting.  I'm not sure whether either interpretation is universally accepted as correct.  My interpretation seems to trivialize the problem, but it is a popular one.  The "epsilon-delta-neighborhood" definition that Steven refers to seems to favor my interpretation as well.  I suggest you ask your teacher (I can't imagine anyone doing this problem for fun - is this homework?) which interpretation they want, and do it the way s/he wants.
EDIT: it might turn out in the end that it doesn't matter which of the two interpretations you use.  But even if this happens, the issue I describe above might be relevant to some other problem, and I suggest you ask your teacher how you should interpret such a question.  
A: For me, it doesn't make sense taking paths that don't belong to the domain, as y=-x, for the point (0,0) should be an accumulation point in order to consider the limit.
I had a great suggestion from a friend.
If you take the limit along $y = (x^4-x^3)^\frac{1}{3}$, you can see that the limit goes to 1 when x goes to 0.
All the other paths I've tried, forced the limit to 0 when x goes to 0.
And this finally proves that the limit doesn't exist, because going through different paths you take different limits!
Along
$$(x,y) = (x,(x^4-x^3)^\frac{1}{3})$$
we have:
$$\lim_{(x,y)\rightarrow (0,0)} \frac{x^2y^2}{x^3+y^3} = \lim_{x\rightarrow 0} \frac{x^2. (x^4-x^3)^{\frac{2}{3}}}{x^3+x^4-x^3}$$
$$ = \lim_{x\rightarrow 0}\frac{x^2(x^3(x-1))^{\frac{2}{3}}}{x^4} = \lim_{x\rightarrow 0}\frac{x^2.x^2.(x-1)^{\frac{2}{3}}}{x^4}$$
$$ = \lim_{x\rightarrow 0}(x-1)^{\frac{2}{3}} = 1$$
A: For the limit to exist, it must exist no matter what direction we take, now the line $y=-x$ as one of these directions, since the limit does not exist (as in it is undefined), in this direction, it cannot for all directions. And thus does not exist.
If you need an $\epsilon-\delta$ style rule, consider an $\epsilon-\delta$ neighbourhood of $(0,0)$
namely $(-\epsilon,\epsilon)\times(-\delta,\delta)$ The line $x=-y$ intercepts this neighbourhood, so the limit is not defined throughout the neighbourhood, and thus does not exist.
A: Note that along the path $y = -x$ the function is not even defined, vs. any other path you choose.
