If the limit of a multivariable function is identical along a certain class of paths, can we claim the existence of the limit? I'm a TA in multivariable calculus course, and I met the following problem. 
If when $(x,y)$ goes to origin along paths $y=mx^k$ and $x=0$ for all $m,k$, $f(x,y)$ along all these paths converges to a same number, can we conclude $\lim_{(x,y) \rightarrow (0,0)} f(x,y)$ exist?
If not, what class of paths where the condition is satisfied, is enough to guarantee the existence of a limit of a function? 
The purpose of this question is to find some sort of the general rule for students to see limit exists without using $\epsilon-\delta$ language or polar coordinate system.  In fact, it's clear that if 
$$\lim_{r \rightarrow 0}f(rcost,rsint)$$converges to the same number uniformly for all $t$, then by the negation of the definition of limit, $$\lim_{(x,y) \rightarrow (0,0)}f(x,y)$$ exists. But this is not what I want, because students cannot understand. 
I really appreciate if anyone can give me any comments or ideas of this question. Thank you very much in advance!
 A: EDITED (Final version).
This example shows that no reasonably simple class of paths suffices to analyze the behavior at a point of a given function $f\colon \mathbb{R}^2\to \mathbb{R}$. Let $\epsilon\colon \mathbb{R}\to \mathbb{R}$ be such that $\epsilon(y)\to 0$ as $y\to 0$ and define 
\begin{equation}
f(x, y)=\begin{cases} 
\displaystyle \frac{\left( \epsilon(y) - x\right)^2}{\epsilon(y)^2 + x^2} , & (x, y)\ne (0,0) \\ 
\displaystyle 1, & (x, y)=(0,0)
\end{cases}
\end{equation}
Any substitution 
$$\begin{array}{ccccc}
x=\gamma(y),& \text{with }&\epsilon=o(\gamma)&\text{or}& \gamma=o(\epsilon)\end{array} $$
produces the limit 
$$
\lim_{y\to 0} f(\gamma(y), y)= 1, $$
so the function is continuous along the path. However, the substitution 
$$
x=m\epsilon(y),\qquad m \in\mathbb{R}
$$
produces the limit 
$$
\lim_{y\to 0} f(m\epsilon(y), y)=1-\frac{2m}{1+m^2}, $$
so the function is almost never continuous along this path (this only happens in the case $m=0$).

Choosing $\epsilon(y)=\exp\left(-y^{-2}\right)$ produces a discontinuous function $f=f(x, y)$ that is continuous along all polynomial paths 
  $$
x=a_1y + a_2 y^2 +\dots + a_ny^n,\qquad n\in\mathbb{N}. $$

This said, the question does admit a positive answer, but of little practical use I am afraid:
Fact. Let $f\colon \mathbb{R}^2\to \mathbb{R}$. The following are equivalent: 


*

*One has that $\displaystyle \lim_{(x, y)\to (0,0)} f(x, y)=L$. 

*For any sequence $(x_n, y_n)\to (0,0)$, one has that $f(x_n, y_n)\to L$. 

*For any continuous curve $\gamma\colon [0, 1]\to\mathbb{R}^2$ such that $\gamma(0)=(0,0)$ one has that $\displaystyle \lim_{t\to 0} f(\gamma(t))=L$. 

*For any smooth curve $\eta \colon [0,1]\to\mathbb{R}^2$ such that $\eta(0)=(0,0)$ one has that $\displaystyle \lim_{t \to 0} f(\eta(t))=L$.
proof. It is clear that $1.\Leftrightarrow 2. \Rightarrow 3.\Rightarrow 4.$ Let us only prove the converse implication $3.\Rightarrow 2.$ Consider a sequence $(x_n, y_n)\to (0,0)$. Define a continuous path $\gamma\colon [0,1]\to \mathbb{R}^2$ as follows: 
$$
\gamma(t) = \Big(  (n+1)(nt-1)x_{n+1} + n[(n+1)t-1]x_n ; (n+1)(nt-1)y_{n+1} + n[(n+1)t -1]y_n\Big)$$
for $t\in\left[\frac{1}{n+1}, \frac1n\right]$. (This is the piecewise linear path with the property that $\gamma\left(\frac1n\right)=(x_n, y_n)$ and $\gamma\left(\frac{1}{n+1}\right)=(x_{n+1}, y_{n+1})$). By assumption, 
$$
f(\gamma(t))\to L,\qquad t\to 0.$$
Therefore, 
$$
f(x_n, y_n)=f\left(\gamma\left(\frac1n\right)\right) \to L,\qquad n\to\infty.$$
  $\square$
Remark. The proof that $4.\Rightarrow 2.$ is the same, with the only added technical difficulty that the path $\gamma$ must be constructed smooth. 
