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 ?