# Epsilon-Delta Proof Structure

I am curious about the Epsilon-Delta proof structure and the best way to do them. I have seen this type of proof done different ways. I would like to accomplish the following from this post:

1. Am I structuring the proof correctly? If not, then what is the correct way?

2. Is my Epsilon-Delta proof correct? If not, then what should I have done instead?

Prove: $\lim_{x\to2} (x^2-4x+5)=1$

Proof: \begin{align*} 0<|x-a|<\delta \implies |f(x)-L|<\epsilon \\&0<|x-2|<\delta \implies |x^2-4x+5-1|<\epsilon \\&= |x^2-4x+4|<\epsilon \\&= |(x-2)^2|<\epsilon \\&= -\epsilon< (x-2)^2<\epsilon \\&= \sqrt-\epsilon< (x-2)<\sqrt\epsilon \\Choose: \delta =\sqrt\epsilon \\&=0<|x-2|<\delta \\&=0<|x-2|<\sqrt\epsilon \\&=-\sqrt\epsilon <x-2 <\sqrt\epsilon \\&=-\epsilon <(x-2)^2 <\epsilon \\&=|(x-2)^2| <\epsilon \\&=|x^2-4x+4| <\epsilon \\&=|x^2-4x+5 - 1| <\epsilon \end{align*} Your help is much appreciated and desired. Thank you in advance.

My recommendation is:

Everything up to the line "Choose: $\delta =\sqrt\epsilon$" should be written on a piece of paper that you don't submit as part of your proof. That piece of paper is merely your "scratch pad". The difference between writing stuff on the "scratch pad" and writing a proof is like the difference between writing down your class schedule and actually attending the classes.

Because of what you have written in the "scratch pad" part of the question, you could be reasonably confident that you'll be able to complete the proof that you do (indeed) write in the rest of the question (setting aside the irregularity in notation).

Because this is an exercise to show that you really understand the $\delta$-$\epsilon$ definition of a limit, however, you should probably use words to describe what is going on in each step. I would certainly not recommend writing "$=$" at the start of each line in the manner you have done.

For example, you can say that for every $\epsilon > 0,$ there exists a $\delta>0,$ namely $\delta=\sqrt\epsilon,$ such that whenever $0<|x-2|<\delta$ it follows that $|x-2|<\sqrt\epsilon$ and therefore ... (and here you continue with the string of facts you have listed, explaining for each fact that it is implied by the one before it).

I think it also wouldn't hurt (and might help) if you state clearly at the end that you have just shown that for any $\epsilon>0,$ there exists a $\delta>0$ (specifically, $\delta=\sqrt\epsilon$) such that $0<|x-2|<\delta$ implies that $|f(x)-1|<\epsilon,$ where $f(x)=x^2-4x+5,$ and therefore $$\lim_{x\to2} x^2-4x+5 = 1.$$

Yes, it sounds a bit repetitious, but I think you're less likely to lose credit for saying correct things too many times than for neglecting to say something that the grader expected you to say.

Regarding the irregular notation, when I see something like

\begin{align*} &=0<|x-2|<\sqrt\epsilon \\&=-\sqrt\epsilon <x-2 <\sqrt\epsilon, \end{align*}

I tend to read it as an assertion of the following string of equations and inequalities: \begin{align*} 0&<|x-2| \\ |x-2|&<\sqrt\epsilon \\ \sqrt\epsilon&=-\sqrt\epsilon \\ -\sqrt\epsilon &<x-2 \\ x-2&<\sqrt\epsilon. \end{align*}

This is for the same reason that people write things like \begin{align*} x^2 - 4x + 5 &= x^2 - 4x + 4 + 1 \\ &= (x - 2)^2 + 1 \end{align*} when what they mean is that $x^2 - 4x + 5 = x^2 - 4x + 4 + 1$ and that $x^2 - 4x + 4 + 1 = (x - 2)^2 + 1.$

Reading your string of statements (equations and inequalities) in this way, most of the statements are true (namely, all the ones whose "left" and "right" both appeared on the same line in the format in which you wrote them), but it clearly is not true that $\sqrt\epsilon = -\sqrt\epsilon$ when $\epsilon > 0.$ So using "$=$" in this way looks very bad.

I recommend using words (instead of trying to use some other mathematical symbol) to show the logical progression of your statements, because it's far to easy for the mathematical symbols to end up being interpreted by someone else in a very different way than you meant them to be.

• How about using 'then' instead of '='? It will be clearer, correct? – Iamlearningmath Sep 24 '17 at 4:03
• I think "then" would work, since in this proof you only need implication from $0<|x-2|<\delta$ to $|x^2-4x+5 - 1| <\epsilon.$ – David K Sep 24 '17 at 11:40

First, you should replace the sequence of $=$ signs with $\equiv$ (or $\Leftrightarrow$). This means for example that the statement $|x^2-4x+4| < \epsilon$ is true if and only if the statement $|(x-2)^2| < \epsilon$. Why is this important? Well, that tells you immediately that you can work backwards from the statement $0 < |x-2| < \sqrt{\epsilon}$ to your desired conclusion. The proof would then look like this. $$|x^2-4x+5-1| < \epsilon \equiv |x^2-4x+4| < \epsilon$$ $$\equiv |(x-2)^2| < \epsilon$$ $$\equiv |x-2| < \sqrt{\epsilon}$$ Then taking $\delta = \sqrt{\epsilon}$, we are done. Note also that you only want bounds for $|x-2|$ and not $x-2$ so some of your lines are not necessary.

• The definition of limit does not require two way implication. Why would someone go for "if and only if" if this is not required? – Paramanand Singh Sep 24 '17 at 3:31
• Because then you don't need to write much once you've found your choice for $\delta$. This would save some time in an exam. It is also a good way to be simultaneously clear (by showing how you obtained $\delta$) and concise. – cofnmarol Sep 24 '17 at 3:56
• But then it will not work everywhere. A one way implication is different from two way implication. In this particular example it was possible to work in this manner. – Paramanand Singh Sep 24 '17 at 4:17