# Limit of $f(x)=x-\lfloor x \rfloor$ $\epsilon-\delta$

For $x\in \mathbb{R}$, let $\lfloor x \rfloor$ denote the largest integer that is less than or equal to $x$. For example, $\lfloor 3 \rfloor=3$ and $\lfloor \pi \rfloor=3$. Define $f:\mathbb{R}\to \mathbb{R}$ by $f(x)=x-\lfloor x \rfloor$. Determine those points at which $f$ has a limit and justify your conclusion.

We assert that $f(x)$ has a limit everywhere but at the integers. If we examine a generic interval $Z=(z,z+1)\subseteq \mathbb{R}$ and look at either endpoint, say $z$ without loss of generality then we can easily see there is no limit there. Take the sequences $z+\frac{1}{n}$ and $z-\frac{1}{n}$. We now consider $z+\frac{1}{n}-\left\lfloor z+\frac{1}{n}\right\rfloor$ which will converge to $0$ where as $z-\frac{1}{n}-\left\lfloor z-\frac{1}{n}\right\rfloor$ will converge to $-1$ Within the interval approaching from either direction will give us a limit value of $0$. If we take $x\in (z,z+1)$, then we observe that $x-\lfloor x \rfloor=0$ for all values of $x$. $\leftarrow$ This is where I am stuck.

For every integer $p$ the map $f$ has no limit at $p$; for, let $p$ be an integer. Then $f(x) \to p - p =0$ as $x \to p+$, but $f(x) \to p - (p-1) = 1$ as $x \to p-$.
We claim that the limit of $f$ at $p$ is $f(p)$ for every non-integer $p$; let $p$ be not an integer. If $x \in \mathbb{R}$, then $|f(x)-f(p)| \leq |x-p| + | \lfloor x \rfloor - \lfloor p \rfloor |;$ taking any $\varepsilon > 0$, if $$|x - p| < \varepsilon' := \min \{ \frac{|p - \lfloor p \rfloor |}{2}, \frac{ | \lceil p \rceil - p|}{2} \},$$ then $|x- p| + | \lfloor x \rfloor - \lfloor p \rfloor | = |x-p|,$ which is $< \varepsilon$ if in addition we have $|x-p| < \varepsilon$; so if $|x-p| < \min \{ \varepsilon', \varepsilon \}$, then $|f(x) - f(p)| < \varepsilon$.
• In the first place we estimate "away" the term $| \lfloor x \rfloor - \lfloor p \rfloor |$. – Benicio Oct 12 '15 at 0:33
Let y is the nearest endpoint to x. That is, If $x \gt \lfloor x \rfloor + \frac12$, let $y = \lfloor x \rfloor + 1$. If $x \le \lfloor x \rfloor + \frac12$, let $y = \lfloor x \rfloor$. Let $\delta = |x - y|$. Then $|x - x_0| \lt \delta \implies | \lfloor x_0 \rfloor - \lfloor x \rfloor | = 0$ $\lt$ any $\epsilon$.