How do limits work with floor/ceiling? I'm interested in the below equation:
$$\frac{n}{\operatorname{floor}(\frac{x}{n})}$$
Plotting with $n = 1..100$ shows the graph being slightly more aliased as $n$ increases and a discontinuity forms from $0..n$. The domain from what I can tell is $(-\infty, 0) \cup [n, \infty)$. I wanted to investigate the limits at each $n$ but not entirely certain how floor and ceiling factor into algebra.
Take for example $n = 2$, 
$$f(x) = \frac{2}{\operatorname{floor}(\frac{x}{2})}$$
How could I find $\lim_{x\to0} f(x)$? Would the limit even exist considering the discontinuity between $[0, 2)$?
I know you can break down the function and help find the limit using the rules of limits:
$$\lim_{x\to b} \frac{p}{q} = \frac{\lim_{x\to b}p}{\lim_{x\to b}q}$$
So more precisely I'm looking for $\frac{2}{\lim_{x\to0}\operatorname{floor}(\frac{x}{2})}$.
For what it's worth, the plot looks like:

I can surmise $\lim_{x\to0^-} f(x) = -2$, am I right in assuming $\lim_{x\to0^+} f(x)$ doesn't exist?
 A: It helps to carefully state definitions.  A workable definition of a limit (of a real function) is something like the following:

Definition:  Let $f$ be a function defined on some domain $D\subseteq \mathbb{R}$ and let $a \in \mathbb{R}$.  Further suppose that there is some $L\in\mathbb{R}$ such that for every $\varepsilon > 0$ there exists some $\delta > 0$ such that if $x \in D$ and $0 < |x-a|<\delta$, then
  $$ |f(x) - L| < \varepsilon. $$
$L$ is said to be the limit of $f(x)$ as $x$ approaches $a$, denoted
  $$ \lim_{x\to a} f(x) = L. $$

I claim that
$$ \lim_{x\to 0} f(x) = \lim_{x\to 0} \frac{2}{\left\lfloor \frac{x}{2} \right\rfloor} = -2. $$
To prove this claim, I have to show that if $\varepsilon$ is any positive number, then I can find a value $\delta$ such that $f(x)$ is within $\varepsilon$ of $-2$ whenever $x$ is in the domain of $f$ and within $\delta$ of zero.
So, let $\varepsilon > 0$ be arbitrary and take $\delta = 1$.  Observe that $f$ is only defined on the set $D = \mathbb{R} \setminus [0,2).$  If $x \in D$ and $|x| < \delta = 1$, then $x \in (-1,0)$, and so $\lfloor x/2 \rfloor = -1$.  But then
$$ |f(x) - (-2)|
= \left| \frac{2}{\left\lfloor \frac{x}{2} \right\rfloor} + 2 \right|
= \left| \frac{2}{-1} + 2 \right|
= 0.
$$
Hence whenever $|x-0| < \delta$ and $x\in D$, we have $|f(x)-(-2)| = 0 < \varepsilon$.  Therefore
$$ \lim_{x\to 0} \frac{2}{\left\lfloor \frac{x}{2} \right\rfloor} = -2, $$
as claimed.

The important point here is that the definition of a limit only cares about what is happening in the domain of the function.  Points where the function is undefined are irrelevant.  As long as we are working with a real-valued function of a real variable, the function of interest is simply not defined on the interval $[0,2)$, and so we don't have to worry about those points.
