Nature of the set of deltas in limit definition

Question

I was proving the fact that if f is differentiable at a, then (cf)’(a)=cf(a)’ After showing it, I started to play with some ideas and wondered from intuition how to see that depending on the value of the constant (magnitude actually), f might be more or less nicely differentiable at that point. To define what I meant by nice , I refer to how close you have to be around a in order to ensure you are “epsilon-good” in the limit definition for the secant slope and f’(a) and cf’(a) respectively. Now, following the principle that in most cases, taking smaller epsilons lead to smaller deltas (unless perhaps you subtract a negligible amount to the next epsilon you choose) I did the following: Suppose f tends to l near a. Now, cf tends to cl near a, but moreover, to be epsilon-good for cf you need to make sure that f is epsilon/|c| good. So, when this is bigger than epsilon (when |c| <1) the deltas to be epsilon-good for cf will be (probably) greater than the ones for f (the opposite happens when |c| >1). Now, to make this idea more solid, I had the following follow up doubt that I would also love to add and make clear: When talking about the set of deltas that ensure you are epsilon-good in any definition of limits in real analysis, call it D, I wonder if this sets are always semi open that is: (0,c] or if sometimes they have to be open : (0,c). I have this question because when making the previous comments, I always assumed that these delta sets had a maximum. I know they have a sup in any case. (open or semi) by the Lu property. However , the thing is whether or not this sup would belong to D. I hope I made everything clear and someone can carefully clear my mind and describe this tiny yet nice remarks

Answer 1

Your question is quite hard to read. I think this sort of answers what you're asking:

Let $U$ be any nonempty set, $a\in U$ fixed. Say $f:U\to\Bbb R$ is some function such that $\lim_{x\to a\\x\in U}f(x)$ exists and equals $L$. Fix $\epsilon>0$.

Define $D:=\{\delta>0:\forall x\in U\cap(a-\delta,a+\delta)\setminus\{a\},\,|f(x)-L|<\epsilon\}$. You want to know if $D$ looks like an open or half open interval.

The answer is that $D=(0,c]$ for some $c$ or $D=(0,\infty)$. Because the limit exists, $D$ is nonempty. Notice that if $\delta\in D$ and $0<\delta'<\delta$ then $\delta'\in D$ too; $D$ is downward closed, so it must in fact be an interval of some kind. If $D$ is unbounded, then $D=(0,\infty)$. Suppose $D$ is bounded.

Let $c=\sup D$. This is the least upper bound; if $0<\delta<c$ then $\delta$ is not an upper bound for $D$, thus there exists some $\delta'\in D$ with $\delta<\delta'$. It follows $\delta\in D$. So, $D$ contains $(0,c)$. If $\delta>c$, then $\delta$ is a strict upper bound of all elements of $D$ so cannot itself be in $D$. The last thing to check is whether or not $c$ is in $D$.

Suppose $x\in U\setminus\{a\}$ has $|x-a|<c$. Then there is some $\delta>0$ with $|x-a|<\delta<c$, and as established $\delta\in D$ must hold and then $|f(x)-L|<\epsilon$ too. Therefore, $c\in D$.

Therefore $D=(0,c]$.