# Question about checking my understanding of the epsilon-delta proof.

My understanding of an epsilon-delta proof is that the purpose is to prove that, regarding the above statement to be proven (e.g. Prove that $$\lim_{x\to2} 3x - 3 = 3$$.), for every $$\epsilon$$ > 0, that there exists some corresponding value of $$\delta$$ > 0 such that for all values of x, the epsilon-delta definition of a limit holds for the above statement, and a main point of the proof is showing that such a value of $$\delta$$ exists, determining that value of $$\delta$$ in terms of $$\epsilon$$ and showing that, for that value of $$\delta$$ in terms of $$\epsilon$$, the above statement fits the epsilon-delta definition of a limit and is therefore true.

• You can avoid "wasting our time" by eliminating unneeded words. I reduced your question by over 50%, saving everyone time, but retaining the key problem. Please try that approach in the future. Apr 6, 2020 at 5:15

You are accurately describing the typical way to prove a limit, though it's worth connecting that explicitly to what the definition is and what the usual way of proving things with quantifiers is.

To start with $$\lim_{x\rightarrow A}f(x) = B$$ is defined to mean:

For all $$\varepsilon > 0$$, there exists some $$\delta >0$$ such that if $$|x-A| < \delta$$ then $$|f(x)-B| < \varepsilon$$.

As with anything in math, people get creative with how to prove these statements, but you can sort of figure out the "normal" way to prove things by applying rules:

First: If you want to prove any statement that looks like

For all $$\varepsilon>0$$...

you start your proof by saying "Let $$\varepsilon > 0$$" and imagining that some hostile being (who wants to see your proof fail) has chosen the worse imaginable value of $$\varepsilon$$ - which you may inspect and work with, but may not change. You must prove the remainder of the statement from this alone - in particular, you have to show for that particular $$\varepsilon$$ that

there exists some $$\delta > 0$$ such that if $$|x- A| < \delta$$ then $$|f(x)-B|<\varepsilon$$.

This is essentially the first thing you list as required.

Okay, so we need to prove a statement that starts with

There exists some $$\delta > 0$$ such that....

The easiest way to do that is to name a particular value of $$\delta$$ that has the desired property, in terms of whatever we know up to that point - which is basically just $$\varepsilon$$. This is basically the second thing you list: we'd like to get a formula that gives us suitable values for $$\delta$$ whenever we have a $$\varepsilon$$. Then, for that $$\delta$$, we must show

If $$|x-A| < \delta$$ then $$|f(x) - B|<\varepsilon$$

which is basically the third requirement you list - of course, we could separate that out as "we are given an arbitrary $$x$$ such that $$|x-A| <\delta$$ and must show $$|f(x)-B|<\varepsilon$$" since this is really just another "for all" statement, but this is usually done implicitly.

If you stick to the rules you listed, you should be fine proving limits and understanding exactly what the challenge is whenever you need to write such proofs. However, it is also worth noting that a limit is exactly what it says it is - if you have some way to show the existence of a suitable $$\delta$$ other than naming one, that's fine. Plus, once you have built a framework around limits, you'll find that you rarely use $$\varepsilon-\delta$$ proofs, but rather work through theorems that do all the dirty work for you.

Here you should consider that this involves the concepts of proximity or approximation.

When one set a positive number $$\varepsilon$$ is to establish an estimate of how near the function, $$f(x)=3x-3$$ is from the limit $$3$$ in terms of how far $$x$$ is from $$2$$. If you need to control the distance from your function to $$3$$ this is the same to say $$|f(x)-3|\le\varepsilon,$$ so, you ought to choose a positive $$\delta$$ to restrict the distance of the variable $$x$$ and estimate $$|x-2|\le\delta$$.

Thus, for example, if you wish to have what is for $$|3x-3-3|\le10^{-3},$$ then you begin by checking $$|3x-6|=10^{-3}$$, which is equivalent to $$|x-2|=0.5\times10^{-3}$$ and, ergo! one gets $$\delta=0.5\times10^{-3}$$. Now, it should be clear that you have $$|3x-6|\le10^{-3}$$, because $$|x-2|\le0.5\times10^{-3}$$: $$\mbox{The variable x should be near to 2 in order to have f(x) be near to 3.}$$