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#2 Re: Help Me ! » Limit Proof Help » 2009-11-06 06:25:36

OK, thank you.  I guess I am just having trouble with the signs when raised to a fractional power, but this helps me to understand a little better.  Watch this space - I'm sure that there will be more questions.

#3 Re: Help Me ! » Limit Proof Help » 2009-11-06 05:56:49

I agree with this also, though I don't know how to arrive at the final answer posted above: 1 / (-M)^1/4.  Can you show me how you would solve?

#4 Re: Help Me ! » Limit Proof Help » 2009-11-06 05:13:07

OK, so if you just use an unsigned M...

As soon as I try to take the square root (twice) of

Don't I have a problem since I can't come up with a real value for a negative number?  If the -1 and M cancel out the negatives, how do I arrive at an answer of

#5 Re: Help Me ! » Limit Proof Help » 2009-11-06 04:15:08

I have tried to solve it this way:

which is different than

#6 Help Me ! » Limit Proof Help » 2009-11-06 04:05:53

Vercingetorix
Replies: 9

I am trying to prove:

,

by using the method where if given any negative number M, we can find a number

such that

satisfies
if
.

So we begin with

and
.

The text gives an answer of

.

I can't quite get this to work out and feel that the signs are giving me some trouble.  Can anyone show me what's happening here?

Cheers.

#7 Re: Help Me ! » The Square Root » 2009-11-05 11:21:28

I stand corrected.  Your teacher is correct, because with a constant like 4 you already know the sign.  When there is a function with an unknown (such as x), it may be positive or negative and hence the ± sign.  Sorry for the bum steer to begin with.

#8 Help Me ! » Inequality Help » 2009-11-05 11:17:38

Vercingetorix
Replies: 2

I am trying to solve

.

The text tells me that the answer is

,

Whereas I continue to arrive at

.

Can someone please shed some light on this?

Cheers.

#9 Re: Help Me ! » The Square Root » 2009-11-05 11:12:48

Hi Anakin,

This is because the square of a negative or positive always results in a positive.

It may help to think of it as:

#10 Re: Help Me ! » Epsilon Delta Proof » 2009-11-03 14:50:53

Awesome - thank you for taking time to help me!

#11 Re: Help Me ! » Epsilon Delta Proof » 2009-11-03 11:10:16

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.

#12 Re: Help Me ! » Epsilon Delta Proof » 2009-11-03 07:11:32

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.

#13 Re: Help Me ! » Epsilon Delta Proof » 2009-11-03 06:43:23

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?

#14 Help Me ! » Urgent Help Requested: Specific Epsilon Delta Proofs » 2009-11-03 01:29:51

Vercingetorix
Replies: 0

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!

#15 Help Me ! » Epsilon Delta Proof » 2009-11-03 01:10:55

Vercingetorix
Replies: 9

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. math_image.aspx?p=SMB02LSMB03x:SMB02FSMB031SMB103SMB02fSMB03,SMB02FSMB031SMB10xSMB02fSMB03SMB02lSMB03SMB013?p=69?p=72

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.  math_image.aspx?p=SMB02LSMB03x:4,SMB02RSMB03xSMB02rSMB03SMB02lSMB03SMB012?p=75?p=38

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!

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