Definition of a limit:
Let $S \subset \mathbb{R^n}$ and let $f$ be a function from $S$ tinto $\mathbb{R^m}$. If $a$ is a limit point of $S$ then a point $v \in \mathbb{R^m}$ is the limit of $f$ at $a$ if $$\forall \epsilon > 0 \exists r > 0 : 0 < ||x - a|| < r \implies ||f(x) - v|| < \epsilon$$ $a,x \in S$.
$f$ is continuous at $a \in S$ if $\lim_{x \to a} f(x) = f(a)$. $f$ is continuous on $S$ if it is continuous at each point $a \in S$.
Claim: any function $\ f: \mathbb{Z} \to \mathbb{R^n}\ $ is continuous.
Let $f: \mathbb{Z} \to \mathbb{R}, f(x) = x$.
If $x, a \in \mathbb{Z}$ such that $|x - a| < 1$, then $|x - a| = 0 \implies x = a$.
Going for a contradiction, if $\epsilon = 1, |f(x) - f(a)| = |x - a|$. We need to find $r > 0$ so that $0 < | x - a| < r$ so that $|x - a| < 1$ to satisfy continuity. But $\forall r < 1, |x - a| < r \implies x = a \implies | x - a | = 0$.
Confusion:
Does this mean that the hypothesis, $0 < |x - a| < r$ is always false for $\epsilon \leq 1$ and thus the statement is a tautology under such conditions? We can certainly get within $\epsilon$ when $\epsilon > 1$, so I don't pay too much attention to that case here.
Alternatively, does this mean that there is no such $r$ to get us within the $epsilon$ bound, therefore false.
I think that the first conclusion is true, but I want to make sure.
Nonetheless, it seems really uncomfortable to have a function
$$ f(x) = \begin{cases} x & \text{ if } x \in \mathbb{Z} \\ \text{0} & otherwise \end{cases}$$
called continuous. Perhaps I don't see the difference between $0$ and "undefined" which is leading to my confusion.
Your first function is continuous because your domain is discrete and inverse image of open sets are open.
Your second function is obviously not continuous at integers ( except zero) because of the jump discontinuities at those points.
You are jumping from zero to an integer and come back to zero. Thus you right and left limits at integers are zero but the value of your function is different.