So I came to this question while trying to answer simpler one: "Is square root function continuous at $0$ or only right continuous?"
If you look at the wikipedia page about $\varepsilon$-$\delta$ definition of the limit: link.
It includes the requirement for the point to which the input approaches to be limit point of the function`s domain.
Now if we go to the wikipedia page about limits of function we get the following definition: link. There the domain of the function itself is restricted to be open interval or real line. So I decided to take examples of definitions from analysis and calculus textbooks. So in Rudin's "Principles of Mathematical analysis" we have the definition which applies to metric spaces but it certainly can be specified to real-valued functions of real variable, so actually his definition is the same as definition on the first wikipage I mentioned.
If we take a look at Spivak's "Calculus", his definition I think can be summarized as follows:
$\forall\epsilon\gt 0 \exists\delta\gt0 \forall x,0\lt|x-a|\lt\delta\to|f(x)-L|\lt\varepsilon$ and also he adds a requirement for $f$ to be defined in some open neighborhood of $a$, except maybe $a$ itself.
And the problem with different definitions is that they are not equivalent. The definition of Rudin and Spivak are not equivalent because e.g. by Rudin's definition square root of $x$ with domain of nonnegative reals is continuous at $0$, but by Spivak's, it's not.
If you look at the definition by Rudin (or wikipage I mentioned first) it allows for limits of functions with domains which are not open intervals, but definition in the second wikipage allows only limits of functions with domains which are open intervals.
So what definition is right (actually there may be some other definitions in other textbooks, so you can give them as an answer to the question) ? Or am I too pedantic about it ?
Best Answer
I don't have either of the books in front of me, but I've had this issue for a long time aswell.
Let me quickly restate the two definitions:
The main quantifier part is the same i.e.:
$$\forall \epsilon\;\exists \delta \;\forall x \;\;((0<|x-c|<\delta)\implies (|f(x)-L|<\epsilon))$$
The difference is that in one case for the limit to exist you require there is some punctured neighborhood of $c$ which is in the domain of $f$ and in the other case you just require that $c$ is a limit point of the domain.
The second definition is a generalization of the first as can readily be seen. Which one is correct is in some sense not a reasonable questions. They are definitions they are both self consistent so they are both "right" as far as definitions go. Rather it is a matter of convention which one is used in a given tradition/school etc. Lately I seem to have seen more of the more restrictive version (with punctured neighborhoods) but personally I seem to remember being taught the more general one.
EDIT On reflection following some insightful comments from @Hurkyl I must admit the choice of wording of "more general one" isn't really great. It implies that the second version is a generalization of the first which probably isn't right in the sense that while you can say a "ring is a generalization of field" the important distinction is I'm calling this a ring. I don't go on calling something without inverses a field.
The thing I wanted to also add is that what I will call the second limit definition has some strange consequences if you let it do work in places like the definition/theorem on continuity. Specifically with that definition of limit you get that $f(x)=x$ with domain $\mathbb{Q}$ is continuous wherever defined. This is not something people generally tend to think of as a continuous function on the reals though.
@Hurkyl made another great point which is that really the trouble is we treat partial functions without the respect they deserve. If you restrict yourself only to functions which are total on $\mathbb{R}$ all the problems go away and the definitions become equivalent.