Is $y = \lim_{n \to 0} \left( x \bmod n \right)$ the same as $y = 0$? My question is simply whether or not
$$
y = \lim_{n \to 0} \left( x \bmod n \right)
$$
is identical to
$$
y = 0.
$$
I don't have a formal education in either number theory or analysis, so I'm not sure if it is possible for a discontinuous function to approach a continuous function.
 A: **Answered here for completeness, so this question does not remain unanswered.
If you define $y=x \pmod n = x-n\big\lfloor \frac{x}{n} \big\rfloor$, then indeed, this function is well-defined for real $x$ and any real positive $n$ [even though convention typically has $n$ to be integral], and $\lim_{n \rightarrow 0}x \pmod n = 0$. Indeed, $x \pmod n$ takes on, for all $x$, the values in $[0, n)$ and does not take on any values larger than $n$ nor less than $0$.
In fact, if you were to graph $x \pmod n$ as a function of $x$, you would get a sawtooth function https://en.wikipedia.org/wiki/Sawtooth_wave [click the link to see the type of graph I am referring to, the function graphed resembles the the teeth of a saw]. For each integral $k$, the function is increasing in $x$ for $x$ in the interval $[\frac{k-1}{n}, \frac{k}{n})$, and on that interval. Then, put informally, the function approaches the point of a "tooth" of the saw [the point $(x=\frac{k}{n},y=n)$] as $x$ approaches $\frac{k}{n}$ from the left, and gets arbitrarily close but does not reach. But then for $x =\frac{k}{n}$ itself, note that $x \pmod n$ is $0$, so the function is back in the deepest part of the valley between the saw teeth, at $(x=\frac{k}{n},y=n)$.
The smaller $n$ is, the finer the sawteeth become i.e., the closer the graph stays to the $x$-axis.
