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The definition in my book is as follows:


Let $f$ be a function defined on an open interval containing $c$ (except possibly at $c$) and let $L$ be a real number. The statement $$\lim_{x \to c} f(x) = L$$

means that for each $\epsilon>0$ there exists a $\delta>0$ such that if $0<|x-c|<\delta$, then $|f(x)-L|<\epsilon$.


With the definition the way it is, I don't see how choosing a smaller and smaller $\epsilon$ implies a smaller and smaller $\delta$.

To me, in order to produce that implication, we would need to restrict $\epsilon$ to be small enough to force $f(x)$ to be strictly increasing/decreasing on $(L-\epsilon, L+\epsilon)$, and define increasing/decreasing without the use of derivatives. However, that is not part of the definition.

P.S. Please refrain from using too much notation for logic, I am not familiar with most of the symbols such as the upside down A and such.

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    $\begingroup$ There's a "for every $x$ such that" in there as well, at least implicitly. I.e., $|f(x)-L| < \varepsilon$ for every $x$ satisfying $0 < |x-c| < \delta$. See what that gives you in your purported example. $\endgroup$ – mrf Oct 12 '15 at 13:42
  • $\begingroup$ It is not necessary; consider $f$ defined on $[0,1]$ as the constant function $f(x)=1$. We can ask the limit of $f$ for $x \to 1/2$ (of course, it is $1$). In this case, we can "squeeze" the $\epsilon$ as we want but we are not forced to decrease the $\delta$. $\endgroup$ – Mauro ALLEGRANZA Oct 12 '15 at 13:44
  • $\begingroup$ Saying ""there exists a δ and as ϵ decreases, so does δ" is not sufficient because it does NOT say that δ goes to 0. And, since the point is to DEFINE "limit", you would have to say precisely what you mean by "goes to 0" without using limits! $\endgroup$ – user247327 Oct 12 '15 at 14:14
  • $\begingroup$ $f(x)=x\chi_\Bbb Q(x)$ is nowhere continuous except at $x=0$, and also nowhere monotonic, and nevertheless $\lim_{x\to0} f(x)$ exists and is $0$. ($\chi_\Bbb Q(x)$ is $1$ when $x$ is rational, and $0$ otherwise). $\endgroup$ – Jean-Claude Arbaut Oct 24 '15 at 10:47
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First of all, there exists functions that every $\delta$ is sufficient for them. For example constant function $$f(x)=2$$

On the other hand there exists functions that force us to select as small $\delta$ as possible. For example take a strictly increasing function $f(x)=x+1$.

function f with delta-epsilon limit interpretation

The greatest $\delta$ we can take is such that $f$ intersects the corners of rectangle created by four lines: $x=L\pm\varepsilon$ and $y=c\pm\delta$. So if $\varepsilon$ shrinks, $\delta$ has to get smaller too.

Now, take a function that is not monotonic, for example $g(x)=2x^2+1$.

function g with delta-epsilon limit interpretation

How small does $\delta$ have to be? Similar situation as above, but now $\delta$ is bounded by upper corners.

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First of all, if a bigger $\delta >0$ is found, you can always find a small one (e.g. $\delta_\epsilon = \min\{\delta, \epsilon\}$) to accompany $\epsilon$, so that it kind of match your intuition that the smaller the $\epsilon$, the smaller the $\delta$.

On the other hand, there are actually good functions around that do not require a smaller $\delta$: for example, if $f$ is a constant function, any $\delta >0$ would suffices no matter how small $\epsilon$ is.

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Let me point out to you that a function like $x \in (0,+\infty) \mapsto x \sin (1/x)$ vanishes infinitely many times in any neighborhood of zero; it is impossible to make it monotonic by restricting its domain. Despite this, $$ 0 \leq \left|x \sin \frac{1}{x} \right| \leq |x| $$ and therefore $\lim_{x \to 0+} x \sin (1/x)=0$.

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The fact that as $\epsilon$ decreases so does $\delta$ generally follows from the behavior of the function. Note that for almost all interesting functions $\delta$ will have to decrease as $\epsilon$ decreases. The only exception are locally constant functions.

As long as the function is not locally constant at $a$ (and has a limit $L$ there) then for every $\chi>0$ there is some $x$ and $\psi>0$ such that $0<|a-x|<\chi$ and $|f(x)-L|>\psi$ (otherwise $\forall x\forall \psi>0$ you have $0<|a-x|<\chi\implies |f(x)-f(a)|<\psi$ so $f$ is constant on $(a-\chi,a+\chi)\setminus\{a\}$). But that means that if you choose $0<\epsilon<\psi$ then $0<\delta<\chi$.

Just to clarify your comment about $x^2$ you can't have your $\delta$ "land you" around $-2$ since $\delta$ bounds the distance from $x=2$. If your delta is big enough to get you all the way to $x=-2$ ($\delta\geq 4$) then unless you $\epsilon>4$ the point $x=0$ (which will be withing $\delta$ of $x=2$) will be too far from the limit $L=4$.

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Topologically speaking, it says that for any vicinity $V(L)$ of $L$ there exists a vicinity $W(c)$ of $c$ such that $f(W(c) \setminus \{c\}) \subseteq V(L)$. From this, topological, perspective, the answer to:

I don't see how choosing a smaller and smaller $ \varepsilon $ implies a smaller and smaller $ \delta $.

It shouldn't, otherwise we are in trouble defining (e.g.) limits when $x \rightarrow +\infty $ where $W(+\infty)$ is something like $(\delta , +\infty)$

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The intuition of a limit is that the closer $x$ gets to $c$, the closer $f(x)$ gets to $L$. The situation isn't symmetrical, it is always possible to find $x$ close to $c$, but maybe not so easy to find $f(x)$ close to $L$.

If you just said, "let me choose $\delta$ and see how small $\epsilon$ can be", you couldn't conclude anything because you'd have no criterion to say that $\epsilon$ is small enough.

So you work the other way, saying "I can make $\epsilon$ as small as I want, and I can still find a $\delta$ that fits".

The exact behavior of the function in the $(\delta,\epsilon)$ neighborhood is irrelevant, it can be as irregular/chaotic/discontinuous as you want, provided it remains bounded. Monotonicity isn't required.

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"With the definition the way it is, I don't see how choosing a smaller and smaller ϵ implies a smaller and smaller δ."

Not necessarily so. It does not have to be monotone.

Delta-epsilon just says: There exists a step in the ongoing appliance of the function which leads to a result nearer to the limit, the distance is smaller than any certain number ϵ. No matter how small this number might be.

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The goal of limit of a function is to describe the behavior (the value) of a function $f$ when we are approaching $c$ (but not equal to $c$)

$\varepsilon$-$\delta$ definition of limit is to define the idea approaching in rigorous sense.

What is the meaning of approaching?

If someone gives you a sufficiently small value $\varepsilon > 0$ and he requires the distance between $L$ and $f(x)$, $x$ is the point you are approaching $L$, is within a distance of $\varepsilon$. (This gives $|f(x) - L| < \varepsilon$)

You can always find another sufficiently small $\delta > 0$ such that if you approach $c$ within a distance of $\delta$, the above requirement is satisfied. (This gives $0 < |x-c| < \delta$)

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