understanding epsilon

7/29/2019 Understanding Epsilon

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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 -delta; proofs.

Lets 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 lets 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


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| 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.

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 | <

then

| f(x) L | <


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Epsilon-delta says:

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

being within units of L.

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

lim f(x) = L

x->c

So, heres 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 units of c,

0 < | x c | <

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

| f(x) L | <

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


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

So heres an example of how this stuff works.

Use epsilon-delta to "show" that

lim 3x - 3 = 12

x->5

I know what youre thinking. Cant we just do this:

lim 3x - 3

x->5

= 15 - 3

= 12

but nooooooo! Thats 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 > 0, there exists a > 0 such that:


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IF

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

0 < | x - 5 | <

THEN

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

| ( 3x - 3 ) - 12 | <

So read that in English as:

IF x is within units of 5 . . .

. . . THEN ( 3x 3 ) is within 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 | < ( x is within units of 5 ), THEN we can conclude that | ( 3x 3 )

12 | < (THEN ( 3x 3 ) is within units of 12 ).

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

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

| ( 3x - 3 ) - 12 | <

| 3x - 15 | <

Lets FACTOR (because we love to factor)

| (3)(x - 5) | <


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3| (x - 5) | <

| (x - 5) | < /3;

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

HMM!!!

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

CHOOSE = /3. (Get used to the idea of CHOOSING )

Then, we go:

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

| x - 5 | <

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 = /3. Youll see that this choice doesnt really hurt the rigor

of what were doing though. . . just keep at it.

Next we have to show that this weve chosen ( = /3 ) WORKS.

So we go back to the original statement:

IF

0 < | x - c | < [ c = 5, chose = /3 ]

0 < | x - 5 | < /3;

THEN

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


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| ( 3x - 3 ) - 12 | <

| 3x - 15 | <

3| x - 5 | <

| x - 5 | < /3

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

IF x is within units of 5 . . . THEN ( 3x 3 ) is within units of 12.

To:

IF x is within /3 units of 5, THEN x is within /3 units of 5.

Which cannot be argued against.

Remember, its not stupid. Its rigorous.

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