Epsilons, Deltas and Limits... Oh My!

Yes, Toto, writing proofs of limits can be as scary as the wicked witch from the east! But do not fear, with the right direction, we can squash those problems with ease.

The first step is to realize that we are proving a limit based on its definition. Suppose we need to prove the statement written in its general form:

lim_{x → a} f(x) = L

Notice that this really is saying that when x is a value close to a, the value of f(x) is close to L. The mathematical statement of this says:

"For any ε > 0, there exists a value δ > 0 so that if 0 < |x-a| < δ then |f(x)-a| < ε."

Now just because the Scarecrow is flapping in the breeze, we don't need to be afraid of this complicated looking formula. Our task is to be able to find a formula for δ in terms of ε so that once you know that the value of x is within δ of the value a (δ says how close), then the value f(x) is within ε of the value L.

To reach that fabled wizard of mathematics called a proof, we just need to follow the yellow brick road outlined below. The proof will always take a form involving four steps corresponding to the four parts of the definition:

1. For any ε > 0: We need to create a proof that works for any ε > 0. So that we have a value to work with, we start with any ε with the requirement ε > 0. So the first statement of the proof is something like, "Given ε > 0" or "Suppose ε > 0" or "Let ε > 0."

2. ... there exists δ > 0 such that: The second step is that we need to provide a recipe for how to provide δ > 0 that will make the rest of the statement true. Unfortunately, by the time we reach this step of the proof, we don't yet know what the right recipe is. Personally, I just write, "Let δ=____" and leave enough space to fill in later.

3. ... if 0 < |x-a| < δ ...: We are starting to prove an implication (if...then... statement). We are successful if we can show the conclusion is true whenever the hypothesis is true. So, to accomplish this, we assume that the hypothesis is true and see what happens. I write, "Assume 0 < |x-a| < δ."

4. ... then |f(x)-L| < ε: This is the conclusion of the implication. And this is also the hardest part of the proof. In the middle of completing the work in this proof, we will discover the recipe needed for δ. At that time, we can go back and fill in the missing pieces.

So the 4th step is the hard one. Don't be cowardly like the lion and give up; there is a method to this as well. For the polynomials that we work with, the value of |f(x)-L| will always factor into something of the form:

|f(x)-L| = |x-a| |"stuff"|

We know from step 3 that |x-a|< δ. We want |"stuff"| to be less or equal to a number, which for convenience in discussion we'll call k. Once we find that number k, then we know:

|f(x)-L| < δ k

So part of our recipe will be to make sure that δ≤ε/k. If the recipe requires no other parts, we can even just use δ=ε/k. With this knowledge, we will have found:

|f(x)-L| < δ k ≤ (ε/k)k = ε

All would be well, except that the wicked professor of the west hasn't yet told you how to find k. Let's step back a moment to the |"stuff"| factor. If you don't, I'll send my winged monkeys to bring you back :-). When we're lucky (and f(x)=mx+b), the factor is already a number. But for any other problem, there will be a formula that still involves x. In these cases, without knowing more about x, we won't know how big the extra "stuff" can become. In order to keep a handle on this "stuff" we are going to require for our recipe that δ itself never gets too large. For the simplest cases, we can require δ ≤ 1. And this means that we can take advantage of knowing that x will be between a-1 and a+1.

In a general problem, we could find the largest value for |"stuff"| using values of x between a-1 and a+1. This might take a bit of work. But for f(x) that is quadratic, "stuff" is going to be another linear looking term. We want that "stuff" to involve x-a, so use x = (x-a) + a.

For example, when a=2, the term x+1 can be rewritten x+1 = (x-2)+2+1 = (x-2)+3. The awesome Triangle Inequality then tells us:

|x+1| = |(x-2)+3| ≤ |x-2| + 3

But we know that |x-2|<δ and we required δ ≤ 1 for our recipe. So |x+1|< 4.

For another example, suppose that f(x)=x²-x, a=3, and L=6. We assume 0<|x-3|<δ and rewrite

|f(x)-L| = |x²-x-6| = |x-3||x+2|

We know |x-3|<δ and we need to find a number k so that |x+2|≤k. Since x is in "stuff", we require δ ≤ 1 and use the triangle inequality:

|x+2| = |(x-3)+3+2| ≤ |(x-3)|+5 < δ+5 ≤6

So |x+2| < 6 (This is our value k=6). Thus we also want to use δ = ε/6 in our recipe. Both requirements are taken care of by the formula δ = min(1, ε/6). So now we know

|x²-x-6| = |x+2||x-3| < 6 δ ≤ 6(ε/6) = ε

We have arrived at the emerald city of our desire and proved the limit statement

lim_{x → 3} x²-x = 6.

But the proof needs to be in the right order:
Given ε>0.
Let δ = min(1, ε/6).
Assume 0 < |x-3| < δ. |x²-x - 6| = |x-3||x+2|
|x+2| = |x-3 + 5|
|x+2| ≤ |x-3| + 5
|x+2| < δ + 5 and δ ≤ 1 so |x+2| < 6
|x²-x-6| = |x-3||x+2| < 6δ and δ ≤ ε/6
So |x²-x-6| < ε
Thus, for all ε>0, there exists δ>0 so that if 0 < |x-3| < δ, then |(x²-x)-6| < ε.
Therefore, lim_{x → 3} x²-x = 6