How do limits work with floor/ceiling?

Question

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?

Answer 1

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.

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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.