When evaluating the limit of $f(x, y)$ as $(x, y)$ approaches $(x_0, y_0)$, we should or should not consider only those $(x, y)$ in the domain of $f(x, y)$ ? I am confused by different practices of Calculus textbooks. Have anyone searched and found some authoritative opinion ?

Thomas Calculus 14e §14.2 Example 2 (Page 802-803) $\lim_{(x, y) \to (0, 0)} \frac{x^2 - x y}{\sqrt{x} - \sqrt{y}}$ considers only those $(x, y)$ in the domain. The authors' answer ($\mathbf{0}$) is the same as the answer by WolframAlpha . See textbook page 802 and textbook page 803 .

Larson Calculus 10e §13.2 Exercise 27 (Page 887) $\lim_{(x, y) \to (0, 0)} \frac{x - y}{\sqrt{x} - \sqrt{y}}$ considers NOT only those $(x, y)$ in the domain. The authors' answer (DNE) is NOT the same as the answer by WolframAlpha ($\mathbf{0}$). See textbook page 887 and solution manual page 1268 .

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    $\begingroup$ Are solutions manuals written by the authors of the textbook? This looks like it's just an error in the solutions manual to me, which isn't uncommon at all. Maybe I'm missing something. $\endgroup$ Commented Aug 5, 2020 at 2:59
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    $\begingroup$ @ChrisCunningham In the (short) answers to odd-numbered exercises at the back part of Larson Calculus 10e the answer to §13.2 Exercise 27 $\lim_{(x, y) \to (0, 0)} \frac{x - y}{\sqrt{x} - \sqrt{y}}$ is DNE. $\endgroup$ Commented Aug 5, 2020 at 3:13
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    $\begingroup$ Every definition of a limit I've seen only considers the domain of the function; see the various definitions on Wikipedia: en.wikipedia.org/wiki/Limit_of_a_function#More_general_subsets. If you don't trust Wikipedia, this is also corroborated by the textbook we used in my Real Analysis class: maa.org/press/maa-reviews/introduction-to-analysis. $\endgroup$
    – Reed Oei
    Commented Aug 5, 2020 at 11:18
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    $\begingroup$ @ReedOei Perhaps you could flesh that out a bit into an answer? I agree with you that in most Analysis texts, the definition will only allow the "test points" to be in the domain of the function. $\endgroup$ Commented Aug 5, 2020 at 11:21
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    $\begingroup$ @ncr $\lim_{x\to 0} \sqrt{x}$ exists, and is equal to zero. There are some elementary texts where they define limits in such a way that this limit doesn't exist, but these definitions are incredibly problematic, and cause problems when students move on to more advanced classes. $\endgroup$
    – Xander Henderson
    Commented Aug 5, 2020 at 14:53

1 Answer 1


Most definitions of a limit of a function only include the domain. See the Wikipedia article about limits or this textbook about Real Analysis (or the various textbooks in this Math StackExchange answer from @twJizhan's comment).

But this is only what the definition is, so I'll try to provide a few (informal) arguments about why the definition should be this way.

  1. Generality: Because we can define limits in such a way that makes more functions have limits at more points in a reasonable manner, we might as well. Of course, we must be conscious of the domain and codomain of functions, but we should always be conscious of them anyway.
  2. Notation: It's quite weird to talk about the behavior of a function outside of it's domain. Every function is undefined outside of its domain; to make a definition like this reasonable, we must have a notion of an "ambient space" that our function lives in. If we define limits in this way, we must not only keep track of the (co)domain of the function (not in the standard limit notation), but also this "ambient space", also not in the notation.
  3. Intuition: An informal understanding of $\lim_{x \to p} f(x) = L$ is that $f(x)$ approximates $L$ arbitrarily well when given inputs "around" $p$. This informal understanding works just fine with only considering the domain of $f$ ("of course" we can only give $f$ inputs in its domain), but falls apart a little if we require that $f$ be defined in some (punctured) open ball around $x$ w.r.t the ambient space—then even a function that does approximate $L$ arbitrarily well, in a well-defined way, will no longer have a limit at $p$.

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