Math 111, Introduction to the Calculus, Fall 2011Help on writing epsilon/delta proofs

The ‘Writing Mathematics’ document has some more general advice so please look at that aswell. But there are some specific things to remember when using the epsilon/delta definitionof a limit.

1. Do the scratch work and formal proof separately and label them as such.

You should turn in the scratch work as well so that I know how you are doing the problem,but please separate it and mark it as scratch work. It is the formal proof that you will getcredit for. If you mess up the proof, I might still give you some partial credit if the scratchwork shows you have some idea of what is going on.

2. The formal proof needs to be explained using words.

It is not enough just to write down a sequence of equations. The epsilon/delta notion ofa proof is a complicated idea and you need to explain what you are doing. This meanswriting in complete sentences. You can of course use equations as part of your sentences,but, especially in these sorts of questions, the equations by themselves are not enough. Seethe examples below to see the difference.

3. Every limit proof has the same structure.

Your proof should always look like this:

Proof. Take any ε > 0. Let δ = ? > 0.

Then, for any x such that0 < |x− a| < δ,

we have...

|f(x)− L| < ε.

You can slightly different words to phrase it, but this is how it should look. Notice that thisstructure reflects the definition of the limit:

limx→a

f(x) = L means

for any ε > 0, there is a δ > 0 such that: for any x with 0 < |x− a| < δ, we have |f(x)− L| < ε

4. Your choice of δ will almost always depend on ε, but is NOT allowed to dependon x.

The idea is that for every ε > 0, there is a δ > 0 that works, but it is usually a different δfor each ε. Therefore, when you pick a δ, it is likely to depend on ε, that is, the formula forδ involves ε.

On the other hand, the values of x that we look at in the proof depend on δ. (They are allthe values in the range a− δ < x < a+ δ. This means it doesn’t make sense for δ to dependon x, because x is chosen after δ.

5. Showing that a limit is not equal to something is similar but sort of reversed.

First you should make sure you understand, and know how to write proofs that a limit ISequal to something.

To show that limx→a f(x) is NOT equal to L, we have to show that it is NOT true that: forany ε > 0, there is a δ > 0 such that: for any x with 0 < |x− a| < δ, we have |f(x)−L| < ε.

Sometimes it is hard to see what you have to do to show something is NOT true. In thiscase, you have to show that:

there IS an ε > 0, such that for ANY δ > 0, there is an x with 0 < |x− a| < δ and |f(x)− L| ≥ ε.

To write up a proof like this, the structure should therefore be:

Proof. Let ε = ? > 0. Then take any δ > 0 and let x = ? .

Then it is true that0 < |x− a| < δ

but, because of...

it is NOT true that|f(x)− L| < ε.

In this proof, you choose a particular number for ε, like 1 or 0.1. You also choose an x, butthis is allowed to depend on both δ and ε.

EXAMPLE 1

Prove that:limx→1

(−2x+ 5) = 3.

BAD ANSWER: (This answer has the right elements but they are not in the right orderand not explained at all.)

Proof.

|(−2x+ 5)− 3| < ε

| − 2x+ 2| < ε

2|x− 1| < ε

|x− 1| < ε/2

GOOD ANSWER:

Scratch work:

|(−2x+ 5)− 3| = | − 2x+ 2| = 2|x− 1| because |x− 1| = |1− x|. (Math sucks.)

2|x− 1| < ε, |x− 1| < ε/2. So δ = ε/2. (I hate calculus.)

(I can write whatever I want here because it’s only scratch work and Professor Ching willnot read it.)

End of scratch work.

Proof. Take any ε > 0 and let δ = ε/2 > 0.

Then, if x is such that0 < |x− 1| < δ

we have|x− 1| < ε/2

and so2|x− 1| < ε

so|2x− 2| < ε

so| − 2x+ 2| < ε

so|(2x− 5)− (−3)| < ε

which is what we needed to show.

EXAMPLE 2

Let f(x) be the function

f(x) =

{1 if x is irrational;

0 if x is rational.

Prove thatlimx→2

f(x) 6= 1.

GOOD ANSWER:

Scratch work:

We want to show that as x gets close to 2, it is NOT true that all values of f(x) get close to1. Well, the problem is that when x is rational and close to 2, the value of f(x) is not closeto 1. In fact, those values are a distance 1 away from 1. So take ε to be less than 1, say 1/2.

Now take any δ > 0. We want to find an x that is within δ of 2 but such that f(x) isnot within 1/2 if 1. So we want x to be a rational number that is within δ of 2. e.g. Ifδ = 0.0000032215, then we could take x = 2.000003, and so on.

To explain this we could just say that there is always a positive rational number smallerthan any given positive number. More explicitly, we could let x be what you get when youtake the first non-zero digit in the decimal expansion of δ and delete the remaining digits,divide by 10 (just in case δ only had the one nonzero digit), then add 2.

(End of scratch work.)

Proof. Let ε = 1/2 and take any δ > 0 Let x be the rational number obtained by

x = 2 +d

10

where d is given by taking the first non-zero digit in the decimal expansion of δ and deletingthe rest.

Then, since d 6= 0, we have x 6= 2, so 0 < |x− 2|. Also, since d ≤ δ, we have

|x− 2| < δ.

Altogether then0 < |x− 2| < δ.

But, because x is rational, we have f(x) = 0 and so

|f(x)− 1| = 1 ≥ ε.