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?

  • $\begingroup$ Perhaps you could have found @JoséCarlosSantos answer if you started from a picture of the graph of the floor function. $\endgroup$ 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$$\frac{\lfloor x\rfloor-\lfloor a\rfloor}{x-a}=0<\varepsilon.$$


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