Limits along what curves suffice to guarantee the existence of a limit? For functions f from $\mathbb{R}^2$ to $\mathbb{R}$, we can define the limit of $f ( x,y)$ as $(x,y)$ goes to $(a,b)$ along the curve $C$ for any continuous curve $C$ passing through (a,b).  And it is a theorem that if the limit of $f(x,y)$ as $(x,y)$ goes to $(a,b)$ along $C$ exists and is the same for all continuous curves $C$ passing through $(a,b)$, then the (unqualified) limit of $f(x,y)$ as $(x,y)$ approaches $(a,b)$ also exists and is equal to the limit along all those curves.
My question is, can we weaken those conditions?  In other words, is there some smaller class of curves passing through $(a,b)$ for which $f(x,y)$ having the same limit for them suffices to guarantee the existence of the limit?  We can't make restrict ourselves to just lines passing through $(a,b)$, as there exist examples of functions which have the same limit along all lines passing through a point, but a different limit along some parabola.  (For instance, $f(x,y) = \frac{x^2y}{x^4+y^2}$ for the point $(0,0)$ and the parabola $y=x^2$.) And I assume that similarly, the limit along any parabola could be the same, but the limit along some cubic function could be different, and that more generally for any $n$, there exists a function which has the same limit for all polynomial curves of degree $n$ or lower, but has a different limit along some polynomial curve of degree $n+1$.  (By a polynomial curve I mean either the graph of a polynomial function, or a curve obtained by rotating the graph of such a function about the point $ (a,b) $.)
Assuming that I'm right about all that, what if we took the class of polynomial curves of all orders?  Then would the function having a limit along those curves suffice to guarantee that the limit exists in general?  Or if that doesn't suffice, what about the class of smooth or analytic curves?  
Any help would be greatly appreciated.
Thank You in Advance.
 A: Let $f$ be the characteristic function of the set $$A:=\left\{\left({1\over n},e^{-n}\right)\>\Biggm|\>n\geq 1\right\}\subset{\mathbb R}^2\ .$$
Lemma. Any  curve
$$t\mapsto\cases{x(t):=t\>p(t)&\cr y(t):=t\>q(t)&\cr }\qquad(t\geq 0)\tag{1}$$
with $p$ and $q$  real analytic functions, $p(0)$ and $q(0)$ not both zero, can contain only finitely many points of $A$. 
Proof. Note that the points of $A$ are lying on the curve
$$\epsilon:\quad y=e^{-1/x}\qquad(x>0)\ .$$
Let $\gamma$ be any curve $(1)$. If $p(0)=0$ then $|y(t)|>|x(t)|$ for all sufficiently small $t$, while $|y|<|x|$ on $\epsilon$. When $p(0)\ne0$ we can eliminate $t$ from $(1)$ and write $\gamma$ as graph of a real analytic function: $y(x)\equiv0$ or
$$ y= x^r(a_0+a_1x+a_2x^2+\ldots)\qquad(0\leq x<\rho)\ ,$$
where $r\geq1$ and $a_0\ne0$. In the second and more dangerous case there is an $h>0$ with
$$|y(x)|>{|a_0|\over 2}x^r>e^{-1/x}\qquad(0<x<h)\ .$$
Therefore the curve $\gamma$ cannot contain any points of $A$ with ${1\over n}<h$.
From this Lemma it follows that
$$\lim_{t\to 0+}f\bigl(x(t),y(t)\bigr)=0$$
for all such curves, even though $f$ is not continuous at $(0,0)$.
