Understanding epsilon-delta proofs

First, get and read this. Really good.

before you even START with epsilon-delta . . .

| x – 3 | < 1;

SAYS in English “x is within 1 unit of 3.”

You have to be fluent with that idea before starting ε-delta; proofs.

Let’s look at a picture of this inequality on the real number line:

So for example, if x = 2.8:

| 2.8 - 3 | < 1
| -0.2 | < 1
0.2 < 1    ===   TRUE

Now let’s try for x = 3.5.

| 3.5 - 3 | < 1
| 0.5 | < 1
0.5 < 1   ===   TRUE

So, | x – 3 | < 1 MEANS x must be within 1 unit of 3 on the real number line.

To be able to understand epsilon-delta at all, you have to get into the habit of looking at an inequality like

| x – 10 | < 5

And immediately just say in a snap “That inequality says that x is within 5 units of 10″

Or
| x – 0.4 | < 0.00001

And say “That inequality says that x is within 0.00001 units of 0.4.”

In general, for any inequality with the format:

| x – c | < a

That says in plain english that “x is within a units of c.”

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Epsilon-delta proofs are actually easy.

The meat of epsilon-delta proofs

The meat of epsilon-delta proofs is just this idea.

if
0 < | x – c | < δ

then
| f(x) – L | < ε

Epsilon-delta says:

AS WE RESTRICT x to being within δ units of c, then, as a result of that restriction, f(x) becomes restricted to being within ε units of L.

If the above statement is true, then and only then can we say

lim   f(x) = L
x->c

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So, here’s the “definition of a limit”, but with more explanation in English words:

The limit:

lim f(x) = L
x->c

exists if and only if

when we restrict x to be within δ units of c,

0 < | x – c | < δ

then, as a consequence of that restriction, we in effect are restricting f(x) to be within ε units of L.

| f(x) – L | < ε

If that happens, then we know that the limit of f(x) as x -> c is equal to L.

Ok, but how do you do an epsilon-delta proof?

So here’s an example of how this stuff works.

Use epsilon-delta to "show" that

lim  3x - 3  =  12
x->5

I know what you’re thinking. “Can’t we just do this:”

lim  3x - 3
x->5

= 15 - 3

= 12

but nooooooo! That’s not good enough for epsilon-stupid. You must “show” it.

Use epsilon-delta to "show" that

lim  3x - 3  =  12
x->5

So, what we use the “definition of a limit” as stated at the top of this page (you should memorize it really for use on tests (YES they DO ALWAYS put epsilon-delta on tests . . )

The limit above exists if and only if for each ε > 0, there exists a δ > 0 such that:

IF
    0 < | x - c | < δ            [ c = 5 though, plug in: ]
    0 < | x - 5 | < δ

THEN
      | f(x) - L | < ε         [ f(x)= 3x - 3, and L = 12, plug in: ]
      | ( 3x - 3 ) - 12 | < ε

So read that in English as:

“IF x is within δ units of 5 . . . “

“. . . THEN ( 3x – 3 ) is within ε units of 12.”

The key to epsilon-delta proofs is you have to relate epsilon and delta.

You have to argue that IF 0 < | x – 5 | < δ ( x is within δ units of 5 ), THEN we can conclude that | ( 3x – 3 ) – 12 | < ε (THEN ( 3x – 3 ) is within ε units of 12 ).

HMM! Hopefully, this is starting to make some sense. Here is how you proceed.

Let’s work with the THEN part (the | f(x) – L | statement), and break it down a bit:

      | ( 3x - 3 ) - 12 | < ε
      | 3x - 15 | < ε

Let’s FACTOR (because we love to factor)

      | (3)(x - 5) | < ε
     3| (x - 5) | < ε
      | (x - 5) | < ε/3;

That looks familiar! Suddenly, the | f(x) – L | < ε looks a lot like the | x – c | < δ statement.

HMM!!!

We can relate the epsilon statement and the delta statement ( 0 < | x – c | < δ ) in this way:

CHOOSE δ = ε/3. (Get used to the idea of “CHOOSING” δ)

Then, we go:

      | x - 5 | < ε/3   [ CHOOSE δ = ε/3 ]

      | x - 5 | < δ

WOW!!!!! How marvellous. It will seem very very strange to you that in the middle of this “mathematical rigor”, we end up going and “choosing” δ = ε/3. You’ll see that this “choice” doesn’t really hurt the “rigor” of what we’re doing though. . . just keep at it.

Next we have to show that this “δ” we’ve chosen ( δ = ε/3 ) “WORKS”.

So we go back to the original statement:

IF
    0 < | x - c | < δ            [ c = 5, chose δ = ε/3 ]
    0 < | x - 5 | < ε/3;

THEN
      | f(x) - L | < ε         [ f(x)= 3x - 3, and L = 12, plug in: ]
      | ( 3x - 3 ) - 12 | < ε
      | 3x - 15 | < ε
     3| x - 5 | < ε
      | x - 5 | < ε/3

Wonder of wonders! It “WORKS”, because the statement has now changed from:

“IF x is within δ units of 5 . . . THEN ( 3x – 3 ) is within ε units of 12.”

To:
“IF x is within ε/3 units of 5, THEN x is within ε/3 units of 5.”

Which cannot be argued against.

Remember, its not stupid. Its “rigorous”.