2
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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
5
$\begingroup$

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$$\frac{\lfloor x\rfloor-\lfloor a\rfloor}{x-a}=0<\varepsilon.$$

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.