TITLE: Is taking limits an operator?
QUESTION [8 upvotes]: Algebraically, is taking limits an operator? If so, unlike many operators, it satisfies many operations like $$\lim_{\dots}(\log(x))=\log(\lim_{\dots} x).$$ I mean, not all operations satisfy this property, so is there an algebraic classification of such operators?

REPLY [8 votes]: The limit is a family of linear functionals on the space of functions which are continuous everywhere (so the limit is well-defined).
Every $x\in\Bbb R$ (or another space) defines a functional $\lim\limits_{t\to x} f(t)$. But because $f$ is continuous this is really just $f(\lim_{t\to x} t)=f(x)$. So those are the usual evaluation functionals.
The idea is that the continuity of the functions assures that this is really a well-defined operator, and that its result is the evaluation functional at the point.
You can talk about functions with certain points of discontinuities, or one-sided limits, but in all of them some weak form of continuity must play a role, or else the limit is not well-defined and it won't be a functional (which is a sort of operator).