# Derivative of floor function using epsilon/delta

For every $$x\in\mathbb{R}$$, let $$[x]$$ denote the floor of $$x$$. I read that the derivative of $$[x]$$ over non-integers is zero, and now I want to show it using epsilon/delta, i.e. show $$0=\lim_{x\rightarrow a}\frac{[x]-[a]}{x-a}$$ for all nonintegers $$a$$.

Let $$\epsilon>0$$. There exists $$\delta>0$$ such that $$0<|x-a|<\delta$$ implies $$|[x]-[a]|<\epsilon$$. If I can get $$|x-a|>1$$, then we are done. But I am not able to make $$|x-a|>1$$. So maybe going through this way is not right. In fact, if I can show that $$\frac{|[x]-[a]|}{|x-a|}\leq C$$ for some constant $$C$$, I'll be done.

What should be my delta?

• Perhaps you could have found @JoséCarlosSantos answer if you started from a picture of the graph of the floor function. Dec 8 '21 at 22:56

Let $$\delta$$ be the smallest of the distances from $$a$$ to $$\lfloor a\rfloor$$ and to $$\lceil a\rceil$$; for instance, if $$a=\frac53$$, then $$\delta=\min\left\{\left|\frac53-1\right|,\left|\frac53-2\right|\right\}=\min\left\{\frac23,\frac13\right\}=\frac13.$$Then$$|x-a|<\delta\implies x\in\left(\lfloor a\rfloor,\lceil a\rceil\right)\implies\lfloor x\rfloor=\lfloor a\rfloor,$$and therefore$$\frac{\lfloor x\rfloor-\lfloor a\rfloor}{x-a}=0<\varepsilon.$$