Do finite-domain functions have limits Suppose we have a function $f: \{1,2,3\} \to \{1\}$ such that $f(x)=1$, the domain of the function is a finite set. In this case, does the limit $\displaystyle \lim_{x\to 2} f(x)$ exist?
 A: You can define limits in very general setting for functions $f:A\to B$, but you then need to assume that $A, B$ are topological spaces. In case you are not thinking in terms of such generalities, but rather have $A, B$ as subsets of $\mathbb {R} $ with the usual topology of $\mathbb{R} $ applicable to $A, B$ then $A$ must be an infinite set and the number $a$ used in $x\to a$ must be a limit point of $A$.

For starters it is much better to ditch the idea of limit points and instead rely on a simpler assumption: in order to define $\lim_{x\to a} f(x) $, the function $f$ must be defined in an open interval containing $a$ with possible exception of point $a$ itself. 
Similarly for defining $\lim_{x\to \infty} f(x) $, the function $f$ must be defined on some interval of type $(a, \infty) $ and now you can figure out the requirement for a proper definition of $\lim_{x\to -\infty} f(x) $. 

For most beginners in calculus assimilating the definition of limit under this assumption is a challenge and it is better to avoid limits in more general contexts which use idea of limit points and metric spaces and topological spaces. There is enough time to study these abstractions if one has understood the concrete version, namely calculus on the real line $\mathbb{R} $. 
A: For the limit you should have a topology on your sets. Then you can take a convergent sequence $(x_n)_n\subset\{1,2,3\}$ with $x_n\to 2$ in your chosen topology. Since on $\{1\}$ just one topology exists and every sequence converge to $1$, especially $(f(x_n))_n$ is convergent to $1$ and you get $f(x_n)\to 1$ which is equal to $\lim_{x\to 2} f(x)=1$.
