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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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