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Vercingetorix

Member

Registered: 2009-11-03

Posts: 21

Epsilon Delta Proof

Someone please help.

I am trying to prove the following two problems using the epsilon delta method: (|f(x) - L| < ∈ if 0 < |x - a| < δ)...

1.

The text arbitrarily allows δ = 1/6 and arrives at an answer of δ = min{1/6, ∈/18}.  I've done the work but can't arrive at this answer, can someone help?

2. 

I don't know how to manipulate |SQRT(x)-2| into the form of |x-4| (< δ) AND arrive at the book's answer; δ = 2∈.

Please help!

Last edited by Vercingetorix (2009-11-03 01:15:57)

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: Epsilon Delta Proof

For question 1, assume we have a variable

.  Our task is to find an expression for

such that

for every

This is equivalent to

Now we just need an expression for δ such that

We go with the 2nd inequality since it is smaller for all positive epsilons.

Edit: My answer is the loosest possible expression for δ.  Obviously the text sacrificed some optimization for the sake of simplicity.  Since you only need to prove that there is some δ > 0 such that |f(x) - L| < ∈ this is fine, you don't need the most optimized expression.

Last edited by TheDude (2009-11-03 04:04:26)

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TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: Epsilon Delta Proof

For 2, do the same thing.

I'll let you finish from there.  Do be careful when working with inequalities like this; you need to be aware of signs.  In this case you can square the inequality because all 3 expression are positive as long as ∈ < 2.  If this is for an assignment, however, I would suggest being explicit about signs when you perform an operation that could cause the direction of the inequality to change, so here you should note that you are assuming ∈ < 2.

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Vercingetorix

Member

Registered: 2009-11-03

Posts: 21

Re: Epsilon Delta Proof

Thank you for helping me, TheDude.

In your solution to the first problem, you mentioned further optimizing

so that it is closer to x.  I am trying to reconcile how my text book arrived at the answer

.

Do you know how the text arrived at this answer?

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: Epsilon Delta Proof

I don't think you quite understood what I meant by 'optimal' answer.  What I meant was that my answer allows the widest range of values for x that still satisfy the requirements of a limit.  However, in order to provide an epsilon-delta limit proof you only need to provide any delta that satisfies the epsilon requirements.  In the case of problem #1, the book's answer of δ = min{1/6, ∈/18} is more restrictive than my answer.  So it is not 'optimized' in the sense that there are values of δ that satisfy the proof that are not covered by their answer, but it is correct in that it satisfies the requirements.

As for how they reached their answer, I'm not sure.  I haven't done an epsilon-delta proof in a while and don't remember how to simplify them.

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Vercingetorix

Member

Registered: 2009-11-03

Posts: 21

Re: Epsilon Delta Proof

Thank you for your help, again.

Can anyone simplify the epsilon-delta proof to show how to reach the answer: δ = min{1/6, ∈/18}?

Cheers.

Avon

Member

Registered: 2007-06-28

Posts: 80

Re: Epsilon Delta Proof

First observe that for all

we have

We know we are going to be able to bound

so the important thing is to bound x away from 0.

If

then

and so

For any

we have:
if

then

Hence if

then

whenever

Vercingetorix

Member

Registered: 2009-11-03

Posts: 21

Re: Epsilon Delta Proof

Thank you, Avon. 

This is how I tried to solve it originally.  Where I get stuck is on the fact that 1/6 < x < 1/2...  Is it that we are choosing 1/6 from that range because it makes the denominator under

the largest possible value, and therefore makes

a smaller value?  Do you believe that there is a reason that the book arbitrarily chose

= 1/6?  When you say to bound x away from 0, is that the reason why?  i.e. an x value that approaches 0 causes the function to increase without bound?

Also, I used this same approach to try to solve the second problem listed in the first post.  The text says that the answer is

.  Does this mean an arbitrary value of 4 was used for

?  If that's the case, do you have any idea why the book wouldn't state the answer as

=

?

Thank you in advance for any insight you can lend to this!

Cheers.

Avon

Member

Registered: 2007-06-28

Posts: 80

Re: Epsilon Delta Proof

You could use any number in the interval (0, 1/3) in the place of 1/6. I believe the book chooses 1/6 because it is the midpoint of this interval.
You are right about why we need to bound x away from 0.

In the second problem we have that

for all

i.e. the domain of the square root function. Note that this function is not unbounded close to 0.

If you define limits for functions only defined on a subset of the real numbers like Wikipedia does, then we only need that

whenever

and

so demanding that

is unnecessary.
Of course you can always make

smaller if you want to.

Vercingetorix

Member

Registered: 2009-11-03

Posts: 21

Re: Epsilon Delta Proof

Awesome - thank you for taking time to help me!