# $\epsilon$ - $\delta$ definition of a limit - smaller $\epsilon$ implies smaller $\delta$?

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.

• 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.
– mrf
Oct 12, 2015 at 13:42
• 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 $\delta$. Oct 12, 2015 at 13:44
• 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! Oct 12, 2015 at 14:14
• $f(x)=x\chi_\Bbb Q(x)$ 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). Oct 24, 2015 at 10:47

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.

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 $f(x)=x+1$.

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 $g(x)=2x^2+1$.

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

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 $\lim_{x \to 0+} x \sin (1/x)=0$.

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 $L=4$.

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 $(\delta , +\infty)$

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.

"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.

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 $0 < |x-c| < \delta$)