I can't comprehend Stewart's explanation for the precise definition of limit and it's driving me nuts

Question

I'm going through a Calculus class for the third time, and I understand everything very basically. I'm able to solve simple limits, derivatives, and integrals, but I can never really comprehend what I'm doing and the logic behind it. I'm on my third Calculus book, and I chose Stewart because I heard it's simpler. I couldn't get past the properties of numbers in the previous ones.

I can (now) understand the entire logic that's going on here. However, I can't understand why he makes some assumptions that allow the logic to happen (not sure if that's the right way to put it). They're not really assumptions, I just don't know why these things appear.

For example, he writes:

Notice that if $0 < |x - 3| < \frac{0.1}{2} = 0.05$, then $|f(x) - 5| = |(2x - 1) - 5| = |2x - 6| = 2|x - 3| < 2(0.05) = 0.1$

How did he know delta is 0.05? How does that prove anything about that bunch of equations and inequalities? When I first read this paragraph I though it meant something like

Notice that if (I assume that delta is half of epsilon) then (this string of equations and inequalities is true)

But a friend explained to me that said string is actually how you arrive at delta. As in, it's a proof of the previous assumption. She rewrote it to me like this and it made sense:

$$|f(x) - 5| < 0.1$$ $$|f(x) - 5| = |2x - 1 - 5| = |2x - 6| = 2|x - 3|$$ $$2|x - 3| < 0.1$$ $$|x - 3| < \frac{0.1}{2} = 0.05$$

Now I'm reading Stewart's statement as

Notice that if (I assume that delta is half of epsilon) then (this string of equations and inequalities proves that I magically made the correct assumption)

I don't know if I'm right to think this statement is contrived. My friend wasn't able to help me past here because she doesn't understand what I'm asking anymore :(

Then I got another question, which is how $|x - 3| < \delta$ and $|x - 3| < 0.05$ being true prove that $\delta = 0.05$. It feels like that merely puts a roof on delta, as if it could be anything below 0.05. What if it's actually 0.005? I believe it's justified by saying that we obtained that number from $\varepsilon = 0.1$, so their relation are proven enough that for $\delta = 0.05$ it'll never result in an out of bounds epsilon. I just can't make the synapses to visualize that.

Answer 1

There are a couple of things going on here. One is the somewhat odd challenge-response nature of the limit definition:

Eloise. The limit as $x$ approaches $3$ of $2x-1$ is $5$.

Abelard. I think it is not.

Eloise. Then you must think that there's some neighborhood of $y = 5$ that I can't force $2x-1$ into by choosing a small enough neighborhood of $x = 3$. Otherwise, $2x-1$ is definitely approaching $5$ in the limit.

Abelard. I guess that's right.

Eloise. Well go ahead then. I challenge you to find such a neighborhood.

Abelard. Fair enough. I challenge you to find a neighborhood of $x = 3$ such that for every $x$ in that neighborhood, $2x-1$ is within $0.1$ of $5$.

Eloise. Sure! I choose the neighborhood $(2.95, 3.05)$.

Abelard. Hmm...I'm not sure that works.

Eloise. You're not sure? Then you must think there's a value of $x$ in that neighborhood for which $2x-1$ isn't within $0.1$ of $5$. I challenge you to come up with such an $x$.

Abelard. [thinks for a moment] I guess I can't come up with such an $x$. OK, I admit that I can't force you outside a $0.1$ margin of $5$. But what about a margin of—

Eloise. Let me stop you right there. Whatever margin you were going to say right then, let me call it $\varepsilon$. Then let me define $\delta = \varepsilon/2$.

Abelard. OK, I'm following.

Eloise. Well then I challenge you to come up with any value that's within $\delta$ of $x = 3$ such that $2x-1$ isn't within $\varepsilon$ of $y = 5$.

Abelard. [thinks again for a longer while] No, I concede again that I cannot. I admit that the limit as $x$ approaches $3$ of $2x-1$ must indeed be $5$.

It's not coincidental, by the way, that I choose Abelard and Eloise as our players in this mathematical back-and-forth (nor that I spell Eloise without her usual H). It harks back to the min-max game semantics of logic formalized by the Finnish logician Jaakko Hintikka, in which the verifier (personified by Eloise and represented by the existential quantifier $\exists$) proposes that there is a limit, a proposition that the falsifier (personified by Abelard and represented by the universal quantifier $\forall$) attempts to puncture by finding a counterexample. For more information, read the Wikipedia plot summary for independence-friendly logic.

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To interpret this definition in a more conventional manner, let me state it in English, but indented, so you can see the scoping:

• No matter what $\varepsilon$-neighborhood of $y = 5$ we choose,
  • there's a $\delta$-neighborhood of $x = 3$, such that
    • for any value of $x$ in that $\delta$-neighborhood*,
      • $2x-1$ is in that $\varepsilon$-neighborhood.

*Other than possibly $x = 3$ itself.

Let's now consider how Eloise might have arrived at her $\delta$. She appears, like Stewart, to have conjured up $\delta = \varepsilon/2$ almost out of thin air, but that's clearly not true. The key is to work backward from $\varepsilon$ toward $\delta$, and then once we get there, we show how to move forward from $\delta$ toward $\varepsilon$. Unfortunately, there is an aesthetic in much mathematical exposition that eschews showing the backward step, so the fact that the forward step works seems almost miraculous.

In fact, it's pretty straightforward, especially in this case. Remember, Eloise's goal is to force $2x-1$ into an $\varepsilon$-neighborhood of $5$, no matter what $\varepsilon$ is. So, she asks herself, how wide an interval around $x = 3$ will still yield a value of $2x-1$ in that neighborhood?

Here, we're aided by the fact that $2x-1$ is strictly increasing. (In fact, of course, it's linear, but never mind that now.) So in order for $2x-1$ not to be too high, we need

$$ 2x-1 < 5+\varepsilon $$

and for $2x-1$ not to be too low, we need

$$ 2x-1 > 5-\varepsilon $$

Adding $1$ to both sides and dividing by $2$, for both inequalities, yields

$$ x < 3+\frac{\varepsilon}{2} \\ x > 3-\frac{\varepsilon}{2} $$

That is, we're OK as long as we don't exceed $3$ by $\varepsilon/2$ or more, and as long as we don't fall short of $3$ by $\varepsilon/2$ or more. More succinctly, as long as we're closer to $3$ than $\delta = \varepsilon/2$, we're good. And this works no matter what value of $\varepsilon$ we're challenged with, so we are done!

That's the process we go through in general, although it might be complicated by some properties of the function in question. For example, our analysis was simplified by the fact that $2x-1$ is monotonic. But if we want to show that

$$ \lim_{x \to 1} x^2 = 1 $$

we have to contend with the fact that $x^2$ isn't monotonic. If we're challenged with $\varepsilon = 2$, we have to make sure that as we decrease $x$ below $-1$, we stay above $-1$ and below $3$. That doesn't turn out to be too difficult, but we can easily conjure up examples where identifying the proper margin isn't trivial. Nevertheless, for exercises in textbooks, we can always do it.

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Finally, as some other answers have pointed out, once we've identified a workable $\delta$-interval, any smaller interval will also work, because that smaller interval will always be a subset of the one we've already shown to work. In our example, if $\delta = \varepsilon/2$ works, then $\delta = \varepsilon/3$ will also work, as will $\delta = \varepsilon/4$, etc. So it's not always necessary to pick the largest possible interval; we just have to find one that works. Sometimes, we choose one that's smaller than strictly necessary, because it makes our demonstration easier. That's completely fine, so long as it's strictly positive; we can't have an interval of width $0$.

Answer 2

You could say that those equalities are both the way to find the appropriate $\delta$, and to actually prove that this $\delta$ works. I'll do the argument for a general $\epsilon$, not necessarily $0.1$, since this doesn't seem to be what's confusing you.

So let $\epsilon>0.$ We want to show that the limit of $f(x) = 2x-1$ as $x \to 3$ is $5$. To prove this, we need to find a $\delta>0$ such that $|f(x) - 5| < \epsilon$ whenever $|x-3| < \delta$.

At this moment you may have no clue which $\delta$ we could try, so let's just rewrite the statement above using the definition of $f(x)$. Note that for any $x$ $$|f(x) -5| = |(2x-1)-5| = |2x-6| = 2|x-3|.$$ So just from writing down the definition of $f(x)$, there's a very simple relationship between $|f(x) - 5|$ and $|x-3|$. From this you should be able to see why $\delta = \epsilon/2$ should work.

To see that it does work, we assume that $|x-3| < \epsilon/2$ and observe that then $$|f(x) -5| = |(2x-1)-5| = |2x-6| = 2|x-3| < 2 \cdot \epsilon/2 = \epsilon.$$

Does this help?

Answer 3

There is no such thing as the value of $\delta$. There can be many different values that work. As long as, whenever we have an $\varepsilon$, we find one value of $\delta$ that works for that $\varepsilon$, we have succeeded. In this case, when $\varepsilon = 0.1$, then $\delta = 0.05$ works. Sure, there are other values that work, but having found this one is enough.