Doubt about the definition of limit in two variables In this discussion Finding $\lim_{(x,y)\to(0,0)}\frac{x^2y}{x^3+y}$ I found that we can consider paths that don't belogs to the domain of $f(x,y)$ to prove that a limit doesn't exist, but my teacher would not agree.
I propose to you the definition of limit that I know.
Let $f:\mbox{dom}(f)\subset\mathbb{R}^2\to\mathbb{R}$ and $(x_0, y_0)$ an accumulation point of $\mbox{dom}(f)$. We say that $$\lim_{(x,y)\to (x_0, y_0)}f(x,y)=L$$ if and only if $\forall\varepsilon>0, \ \exists\delta>0$ such that if $(x,y)\in \left(B_{\delta}(x_0,y_0)-\{(x_0,y_0)\}\right)\cap\mbox{dom}(f)$ than $|f(x,y)-L|<\varepsilon$
To show that a limit doesn't exist, I have to find two path $P_1(x,y), P_2(x,y)$ such that $$P_1(x,y), \ P_2(x,y)\in\mbox{dom}(f)\ \ \ \mbox{locally}$$ and $$\lim_{(x,y)\to (x_0, y_0)}P_1(x,y)=(x_0, y_0)\wedge \lim_{(x,y)\to (x_0, y_0)}P_2(x,y)=(x_0, y_0)$$ but $$\lim_{(x,y)\to (x_0, y_0)}f(P_1(x,y))=\ell_1\wedge \lim_{(x,y)\to (x_0,y_0)}f(P_2(x,y))=\ell_2$$ with $\ell_1\ne \ell_2$.
In the discussion that i linked, I discovered that I can choose all possible path... but this is strange to me, and I'm now confused. Please help me to understand. Thank you.
 A: Addendum: To clarify, the one-sided limit definition I use below is based on the only limit definition I've ever seen at the calculus level, which is where I believe this discussion should be aimed since your question looks very much like a question about limits in a standard Calculus 3 course.  As discussed in the comments, it is true that in higher math a slightly different definition may be used in more generality.  It seems that you learned a Calculus 3 version of this more general definition, which is fine.  So what it really comes down to is you're working with a different definition, and under this definition your original notions are correct.
Original answer:

I left a comment to you on that discussion you linked.  I'll expand on it here.
In short:  A path not being in the domain of $f$ does not mean the path can be excluded from consideration in the limit.  The same is true in the one-dimensional case as well.  When we want $\displaystyle \lim_{(x,y) \to (a,b)} f(x,y)$, we want it from all paths.  Just like how when we want $\displaystyle \lim_{x\to c} f(x)$, we want it from all paths.
Consider the one-dimensional case:

In one dimension, when we say that $\displaystyle \lim_{x\to c} f(x)$ exists, it means the limit exists as $x$ approaches $c$ from all directions.  Thus if there is one single direction where the limit doesn't exist, then the entire limit itself (from all directions) doesn't exist.  In the one-dimensional case, there are only two single directions:  from the left $(x \to c^-)$ and from the right $(x\to c^+)$.

For example, $\displaystyle \lim_{x\to 0} \sqrt x$ does not exist, because for $f(x) = \sqrt x$, we can't have $x \to 0$ on the path $x < 0$ (i.e., we can't have $x \to 0^-$) because this path is not in the domain of $f$.
$\bigg[$Side note: We do have the one-sided limit $\displaystyle \lim_{x\to 0^+} \sqrt x = 0$, because $x\to 0^+$ is fine for $f(x) = \sqrt x$.$\bigg]$
Similarly:

In two dimensions, when we say that $\displaystyle \lim_{(x,y) \to (a,b)} f(x,y)$ exists, it means that the limit exists as $(x,y)$ approaches $(a,b)$ from all directions.  Thus if there is one single direction where the limit doesn't exist, then the entire limit itself (from all directions) doesn't exist.  In the two-dimensional case, there are infinitely many directions.

For example, $\displaystyle \lim_{(x,y) \to (0,0)} \frac{x^2y}{x^3+y}$ does not exist, because for $f(x,y) = \dfrac{x^2y}{x^3+y}$, we can't have $(x,y) \to (0,0)$ on the path $y=-x^3$ because this path is not in the domain of $f$.
A: You are correct: most authors would agree that you need to consider the paths that belong to the domain of $f$. Let's try to understand why.
Consider the identity function
$$id: \{0\} \to \{0\}$$
It is clear that it should be continuous and the limit at $0$ exists and equals $0$ . But if you consider $0$ as an element of $\mathbb{R}$, then if you could take any path to $0$ (no matter if it was defined or undefined), then we could take the sequence along $\frac{1}{n}$ and since that path is undefined, the limit wouldn't exist, which does not make any sense! Thus, we clearly see that we need to consider the paths that belong to the domain of $f$.
As an example, the limit of $f(x)=\sqrt{x}$ as $x \to 0$ exists and equals $0$. Anyone claiming otherwise is probably using some simplified version of the definition of the limit (probably used in introductory classes like pre-calculus such as the limit exists iff two-sided limits exist and are equal) that is not fully correct.
