TITLE: How "messy" can a multivariable function be?
QUESTION [22 upvotes]: In calculus lectures, we are told that approaching a limit in a multivariable function through some paths does not guarantee the existence of the limit, for example: here. And one can see that, in the following example its easy to see that there are $2$ paths that seem to point, in one case, that the limit is $0$ and in another case, that the limit is $1$ and hence it does not exist. 
$$f(x,y)=\frac{2xy}{x^2+y^2} \quad f(0,0)=0  $$
But the objects we have in question are continuous functions with the exception of perhaps, the point $(0,0)$ or some other strategically placed point. It seems intuitively reasonable that one can approach $(0,0)$ though $2$ or some "low" number of different paths and have different possible limits but it doesn't seems reasonable that one could have a function such that the possible limit is different for $278$ paths, for example. 
I'm not sure if these lectures try to take into account the full generality of multivariable functions, but  it seems absurd that a  $2-$variable function can have such a messy structure that actually allows it to have an arbitrary number of paths that suggest (each one) different possible limits. But I don't know what to say/think when the number of variables is greater than $3$.
EDIT: There are interesting related sub-questions I forgot to add:

Is it possible that we test the existence of a limit with all the lines and polar coordinates, obtain a possible limit $L$ for these tests and yet, the function have another path such that the possible limit for this path is different of $L$? 
The question above glooms to this: Isn't there a minimum number of paths given by families of curves in which, the possible limit is $L$ for all of them and this actually guarantees that the limit is $L$? I make this question because it seems extremely counter-intuitive that given - for example - tests with all the lines, all the parabolas and polar coordinates all with possible limit $L$, there could still be one path that gives a possible limit different of $L$.

REPLY [8 votes]: Here's an interesting exercise related to your question: Suppose you have a function $f(x,y)$. If it is continuous at the origin along every (continuous) path passing through the origin, then in fact it must be continuous (as a function of two variables) at the origin. In other words, if a function fails to be continuous at the origin, you can construct a (continuous) path through the origin along which the function fails to be continuous.