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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.

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  • $\begingroup$ Not only you "can", but rather, imo, you must check all possible paths or, what is the same, any path in some (open) neighborhood of the limit point. $\endgroup$ – DonAntonio May 22 '17 at 17:29
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    $\begingroup$ Aside remark: While considering paths is sufficient in nice spaces, it is not in general. Better stick to the definition, which uses neighbourhoods and not paths. $\endgroup$ – Daniel Fischer May 22 '17 at 18:32
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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.

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  • $\begingroup$ thank you so much for your answer. Now I am more confident in the definitions I know... but now I don't understand why there are many definitions of limit. I though that math is the same all over the World. :| $\endgroup$ – Ixion May 22 '17 at 20:52
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    $\begingroup$ People define things the way they want. Also, definitions change over time. For example, if you are familiar with compactness, the definition of compactness used to be different than the currently used one (people used to take as a definition of compactness what we call today limit point compactness). There is no one person/book that can claim "limits are defined in this way and everything else is wrong." Who would have an authority to say so? The higher in math you'll go, the more you'll realize how mathematicians oftentimes argue over what the definitions of certain concepts should be. $\endgroup$ – Pawel May 23 '17 at 6:51
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    $\begingroup$ When it comes to basic concepts that have been used for centuries (like limits), there is usually one commonly used definition that everyone accepts. So concepts such as continuity, limits, derivatives, topology etc are "the same all over the world," which means that 99% of mathematicians use the same definition for these concepts. $\endgroup$ – Pawel May 23 '17 at 6:54
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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 $f$.

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  • $\begingroup$ Nice explanation...but this was long discussed in the past comment. I think the OP's doubt is what is the actual definition of limit? This, of course, can depend on a particular author's definition, but as you mention, and was already mentioned before, trying to generalize the definition and spirit of the definition in one variable, we must, imo, assume the same in several variables. $\endgroup$ – DonAntonio May 22 '17 at 18:08
  • $\begingroup$ I have to object. $\lim\limits_{x\to 0} \sqrt{x}$ does exist (and equals $0$) according to the definitions of a functional limit that I consider useful. $\endgroup$ – Daniel Fischer May 22 '17 at 18:20
  • $\begingroup$ @DanielFischer, ok, but this discussion is at the introductory calculus level, and at that level we say $\displaystyle \lim_{x \to c} f(x)$ exists and equals $L$ if and only if $\displaystyle \lim_{x\to c^-} f(x)$ and $\displaystyle \lim_{x\to c^+} f(x)$ both exist and both equal $L$. I've not seen a different (non-equivalent) definition of $\displaystyle \lim_{x\to c} f(x)$ at this level. $\endgroup$ – tilper May 22 '17 at 18:21
  • $\begingroup$ The more I hear about calculus, the more it looks like a terrible idea. The OP here has learned a standard version of limit that generalises well to arbitrary topological spaces. The existence and equality of the one-sided limits doesn't even generalise to $\mathbb{R}^n$. And it leads to learning things like "$\lim\limits_{x\to 0} \sqrt{x}$ doesn't exist" that then must be unlearned when one starts doing analysis or topology. $\endgroup$ – Daniel Fischer May 22 '17 at 18:26
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    $\begingroup$ @Ixion, yes, that's more or less what I was getting at in my addendum on my post. It seems that you're just working with a different definition of a limit. $\endgroup$ – tilper May 22 '17 at 23:33

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