Does the definition of limit correspond to the intuitive notion? [closed]

Question

I have been pondering the question of whether the formal definition of limit captures well our intuitive notion of it now for the past few days, with no headway at all. Perhaps I could find some resolutions here.

It is well known that if for every $\epsilon>0$ there is some $\delta>0$ such that whenever $|x-x_0|<\delta$ then $|f(x)-L|<\epsilon$ then we say that $\lim_{x\to x_0}f(x)=L.$

My question concerns whether this definition captures the intuitive idea of $f(x)$ approaching $L$ as $x$ approaches $x_0.$ Precisely,

(i) Although it is not explicit in the formal definition that $\delta$ is arbitrary, but can it be shown that whenever $\delta$ exists according to the definition then it is arbitrary? In particular can it be shown that there is no lower (positive) bound on $\delta$ for every $\epsilon>0$?

(ii) Apart from this, can it be shown that $\epsilon$ becomes arbitrarily small as $\delta$ becomes arbitrarily small (obviously, this depends upon the truth of the claim in (i) above)? That is, is the dependence of $\delta$ on $\epsilon$ (or vice versa) such that $\epsilon\to 0$ as $\delta\to 0,$ for want of a better way of stating this? In fact, this would seem to be the original problem the former definition was designed to rectify, namely to make these intuitions of "arbitrarily getting near to" precise. In that case, has this been accomplished if it is not explicit from the definition that it accurately captures the intuition? Or, despite being able to frame the question otherwise, can it indeed be shown that the tendency of $\delta$ to $0$ forces $\epsilon$ to $0$? Otherwise what shall we say?

Many thanks.

Answer 1

The answer to your question is that Yes, it captures the intuitive understanding of $f(x)\to L$ if $x\to x_0$ (except that technically one requires $0<|x-x_0|<\delta$ rather than $|x-x_0|<\delta$ unless you’re expecting $f$ to be possibly continuous at $x_0$); however, for the enumerated questions (which probably you seek answers to better understand your main question),

(i) No, $\delta$ is not arbitrary but the supremum of all possible values of $\delta$ is dependent on $\epsilon$; the statement “such that whenever $0<|x-x_0|<\delta$” simply means that $|x-x_0|$ can be arbitrarily smaller than $\delta$——and I suppose this is what you mean when you ask “... $\delta$ is arbitrary?” The very construct of the definition does not demand any lower positive bound on $\delta$——in fact, the infimum or all possible values of $\delta$ is $0$ itself.

(ii) No, one does not need to show that $\epsilon$ becomes arbitrarily small as $\delta$ becomes arbitrarily small, and it can be to the contrary even in some case. To better see this, take the function $$f(x)= \left\lbrace \begin{array}{ll} 0&\mbox{if $x\ne 0$}\\ 1&\mbox{if $x=0$} \end{array} \right. $$ It is immediately clear that “for every $\epsilon>0$ and any $\delta>0$, we have that $|f(x)-0|<\epsilon$ whenever $0<|x-0|<\delta$”; thus, by the definition, we have that $\lim_{x\to 0}f(x)=0$. Clearly, in this particular case, any positively-valued $\delta$ (as a function of $\epsilon$) works——that is, for each $\epsilon$, the supremum of all ‘satisfiable’ values of $\delta$ is $\infty$; thus, if you take $\delta:=2^{-\epsilon}$ say, then obviously $\epsilon\to\infty$ as $\delta\to 0$.