# Questions about the epsilon-delta definition of a limit

The definition of a limit I am working with:

Let $f(x)$ be a function defined on an open interval around $a$ (and $f(a)$ need not be defined). We say that the limit of $f(x)$ as $x$ approaches $a$ is $L$, i.e.

$$\lim_{x \to a} f(x) = L$$

if for every $\epsilon > 0$ there exists a $\delta > 0$ such that for all $x$ $$0 < |x - a| < \delta \implies |f(x) - L| < \epsilon$$

1. First off, is this definition accurate? I assume open interval just means something of the form $(a, b)$ that doesn't include endpoints, but don't we also need that interval to be continuous? No discontinuities or asymptotes or jumps or shifted points or anything like that? Because in theory we could make $\epsilon > 0$ large enough to where it would cover undefined or discontinuous parts of the function.

For example If we pick a really big epsilon then we'd be covering an area that has a discontinuity. Or would we say that if the limit exists we should also be able to find a delta so small that it bypasses all the discontinuities once we get close enough to the point we want? I don't know if we need to add anything more to the definition for this to work though.

1. I am confused about the usage of "for all". Like do they mean literally for all $x$? As in all $x \in \mathbb{R}$? All $x$ in $(-\infty, \infty)$? Or only the $x$ for which $0 < |x - a| < \delta$ is satisfied? It's not clear to me how "all" is referencing $x$ or how it's typically used.
• 1. There's a qualifier "for every" in front of $\epsilon$, so in practice, you'll set $0 < \epsilon$ (< any number). You can set $\epsilon$ small enough (say, smaller than the jump height) to get rid of the discontinuity. 2. Consider [Dirichlet's function[(en.wikipedia.org/wiki/…), which evaluates every rational number to one, and zero otherwise. – GNUSupporter 8964民主女神 地下教會 Feb 26 '18 at 20:06

The definition is correct. Note that making $\epsilon$ larger makes the condition $$0 < |x - a| < \delta \implies |f(x) - L| < \epsilon$$ easier to satisfy. In other words, if you find a $\delta$ that works for some $\epsilon$, it also works for every larger $\epsilon$. So the real content of the definition is that no matter how small $\epsilon$ is, you can still find a $\delta$ that works. In other words, no matter how close you want to force $f(x)$ to be to $L$, you can find a small $\delta$ such that when $x$ is within $\delta$ of $a$, $f(x)$ is that close to $L$.

"For all $x$" literally means "for all $x$". That is, $x$ can be anything, not even necessarily a number! It's not necessary to mention $x$ being a number or to require $0<|x-a|<\delta$ in this part, since if that inequality is not true, then the implication $$0 < |x - a| < \delta \implies |f(x) - L| < \epsilon$$ is vacuously true (a false statement implies anything). In other words, it's saying that no matter what $x$ is, if $x$ happens to be a real number such that $0<|x-a|<\delta$, then $|f(x)-L|<\epsilon$ is true.

(To be clear, for this to make sense, you must interpret $|f(x)-L|<\epsilon$ as being false if $x$ is not a real number in the domain of $f$ and similarly $0 < |x - a| < \delta$ as being false if $x$ is not a real number. That way you can assign a truth value to the implication no matter what $x$ is.)

• If $f(x)$ is defined on an open interval containing $a$, then wouldn't all the $x$'s needed be in the domain already? – Adrian Keister Feb 26 '18 at 20:10
• Perhaps, but the implication works like this: $0<|x-a|<\delta$ implies $|f(x)-L|<\epsilon$. The first inequality already forces $x\in(a-\delta,a)\cup(a,a+\delta)$. – Adrian Keister Feb 26 '18 at 20:15
• So "if p then q" is true even if p is false? That seems strange / possibly problematic? Why do we claim something is true without having the evidence to support it? – user535424 Feb 26 '18 at 20:51
• "If p then q", as a whole, is always true when p is false. These are the "trivially true" cases. This says nothing about whether q is true. We mean the implication as a whole is true when p is false. – Adrian Keister Feb 26 '18 at 20:54
• "If the moon is made of cheese then 1+1=3" is true? – user535424 Feb 26 '18 at 21:21
1. The definition is accurate. All intervals are by definition contiguous, if that's what you mean by "continuous". Functions are continuous, but continuity is not defined for an interval.

2. Here's the thing about limits: you simply aren't concerned about the behavior of the function far away from the point $a$: you are ONLY concerned about the behavior of $f(x)$ NEAR $a$. How near? As close as anyone (not you) wants. In other words, limits ask the question, What happens when you keep on zooming in, and zooming in, and zooming in, ... But you don't actually arrive at $a$. So you wouldn't pick a big epsilon. Epsilon is always thought of as a very small number. For that matter, so is delta.

You are asking all the right questions!

Let me give you an extreme example. Then I'll answer you question.

Let $$f(x)$$ then the function that.

If $$x$$ is irrational then $$f(x) = 2+x$$.

If $$x = \frac ab$$ where $$\frac ab$$ is a rational number in "lowest terms", then $$f(x) = 2 + \frac 1{b}$$. (We'll assume the denominator is not negative although the numerator might be.)

Then $$f(0)$$ is undefined (as $$2 + \frac 1n)$$ is undefined) and this weird function is certainly not continuous. I won't (at this time) go into the exact meaning of contininuous but you can see it jumps from $$2$$ to $$2 \frac 1{b}$$ as $$x$$ goes from rational to irational values and as rationals with low denominators or infinitely close to rationals with high denominators, it jumps about like a flea.

Claim 1: $$\lim_{x\to 0} f(x) = 2$$.

Claim 2: $$\lim_{x \to 1}f(x) \ne 3$$

Claim 3: $$\lim_{x \to 1}f(x) = k$$ will be false no matter what $$k$$ we pick.

To prove that $$\lim_{x\to 0} f(x) = 2$$ we want to show that we can "force" the $$f(x)$$s to get "very close to $$2$$" if we force the $$x$$s to get "very close" to $$0$$.

If we want the $$f(x)$$s to be within $$\frac 14$$ of $$2$$ we can force the $$x$$s to be within a $$\frac 14$$ of $$0$$. If $$|x| < \frac 14$$ then either $$x$$ is irrational or $$x =\frac ab$$ for some $$a,b$$.

If $$x$$ is irrational, then $$f(x) = 2+x$$. $$x$$ is within $$\frac 14$$ of zero, so $$f(x)$$ is within $$\frac 14$$ of $$2$$.

If $$x= 0$$ we have a problem in that $$f(x)$$ undefined. Well, that's not really a problem because we are interested in the behavior of $$f(x)$$ near $$0$$. Not at $$0$$.

If $$x = \frac ab$$ is rational and $$x \ne 0$$ and $$|x| < \frac 14$$ then $$b \ge 4$$ and $$\frac 1b < \frac 14$$. That means $$f(x) = 2 + \frac 1b$$ is such that $$2 < f(x) < 2 + \frac 14$$ so $$f(x)$$ is with $$\frac 14$$ of $$2$$.

So we forced $$f(x)$$ to be within $$\frac 14$$ of $$2$$ by forcing $$x$$ to be within $$\frac 14$$ of $$0$$.

That was one example, can we come up with something for all choices of how close we want to be.

If we want $$f(x)$$ to be within $$\epsilon$$ of $$2$$ for any very very small value of $$\epsilon$$, we can do this by forcing $$x$$ to be within $$\delta = \epsilon$$ of $$0$$.

If $$x$$ is irrational, $$f(x) = 2+x$$. But $$x$$ is within $$\epsilon$$ of $$0$$ so $$f(x)$$ is within epsilon of $$2$$.

If $$x$$ is rational and $$x = \frac ab; a\ne 0$$ and $$|\frac ab | < \delta = \epsilon$$. Then $$b > \frac 1{\epsilon}$$. SO $$f(x) = 2 + \frac 1b < 2 + \epsilon$$. So $$f(x)$$ is within $$\epsilon$$ of $$2$$.

So that does it. We can force $$f(x)$$ to be as close to $$2$$ as we like by forcing $$x$$ to be within a certain distance of $$0$$.

Claim 2: Trying to force $$f(x)$$ to be close to $$3$$ by forcing $$x$$ to be close to $$1$$.

Let's try to force $$f(x)$$ to be within $$\frac 14$$ of $$3$$.

Well matter how close we chose $$x$$ to be $$1$$ we will find a rational, $$r$$ with a denominator more than $$4$$ and $$f(x) = 2 +\frac 14 < 2+\frac 14$$ and that will not be within $$\frac 14$$ of $$3$$.

Indeed for any $$x < 1$$ or $$x > 1$$ that we pick, we will find we can always find that there is an integer $$N > \frac 1{|x-1|}$$ and either $$x < 1 - \frac 1N$$ or $$1+ \frac 1N and so $$f(1 \pm \frac 1N ) = 2 + \frac 1N$$ but there will also be an irration $$y$$ so that $$x < y < 1$$ or $$1 < y < x$$. and $$f(x)$$ is within $$|x-1|$$ of $$3$$.

So we can't force $$f(x)$$ to get close to anything at $$x=1$$.

....

Hopefully that makes the definition clearer.

$$\lim_{x\to a} f(x) = L$$ means

For any distance $$\epsilon > 0$$, no matter how small, we can force $$f(x)$$ to be within $$\epsilon$$ of $$L$$ (i.e. force $$|f(x) - L | <\epsilon$$), by finding a $$\delta$$ so that whenever we have $$x$$ within $$\delta$$ of $$a$$ (i.e. $$|x - a| < \delta$$) we will have to have $$|f(x) - L | < \epsilon$$.

So on to your questions:

1) " I assume open interval just means something of the form (a,b) that doesn't include endpoints, but don't we also need that interval to be continuous?"

The open interval is around the inputs near $$a$$. It has nothing to do with the outputs $$f(x)$$, which can hop about like fleas (provided the hop in increasingly smaller hop for $$x$$s near $$a$$.

So all this is saying is there's a small area around $$a$$ where all the $$x$$ around $$a$$ will have $$f(x)$$ well-defined.

2) "I am confused about the usage of "for all". Like do they mean literally for all x?

No. They mean all $$x$$ so that $$|x - a| < \delta$$. In other words they mean all the $$x \in (a-\delta,a+\delta)$$.

• The usage of "for all $x$" here seems to contradict the other answer by Eric -- I am not sure which to believe is right! – user535424 Feb 27 '18 at 6:57
• You are right that Eric and I interpret the "for all" differently. But we both interpret it in ways that are equivalent. "For all x, if x does K then...." has the exact same results as "For all x where x does K then...". In Erics we let x be anything but we only look at what happens when x does K. In my case we only look at the x where x does K. We both end up looking at the same thing. I think language and explanation wise, mine is clearer. However on rereading, I think Eric is literally more correct to how the definition was worded. – fleablood Feb 27 '18 at 17:20