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I need some help with understanding this:
[url=http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preclimsoldirectory/PrecLimSol.html#SOLUTION%204]http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preclimsoldirectory/PrecLimSol.html#SOLUTION%204[/url]
Refer to the part where we arbitrarily assume d<=1. I don't understand this part. I know d is an arbitrary number, but will it affect the results if I let the range of d be anything else other than 1? What if I choose d to be 100?
This question / problem has been solved by stonebroimage
hey m8 , i'm a design engineer so i could help you with this. what part do you want to elaborate on? i don't see d at first glanse.
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CyPhErIoN: hey m8 , i'm a design engineer so i could help you with this. what part do you want to elaborate on? i don't see d at first glanse.

Sorry I meant delta instead of 'd'.
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CyPhErIoN: hey m8 , i'm a design engineer so i could help you with this. what part do you want to elaborate on? i don't see d at first glanse.
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lowyhong: Sorry I meant delta instead of 'd'.

Lot of text there..would be helpful if you could direct us to the lines you want? for instance, copy part of the line that is closest to the problem you're looking at so we can do a Search for it
It's been aaages since I did basic calculus like this now but I'll give it a go ...
The value of delta required to make the proof legit will depend upon the value chosen for e (epsilon).
The key is that once you find a value for delta that works, all smaller values of delta will also work.
How to first choose delta is arbitrary. Some insight into the problem (for example from a graph) may provide a good initial guess. You then test this value of delta and see if the assumption (as per the definition) holds. It is entirely possible to guess a value for delta for which the expressions do not hold. In that case, you must simply guess again (iterate on delta).
Hence, delta <= 1 is equally as arbitrary, and valid, as an initial guess, as delta <= 100 or delta = 141.
Post edited August 27, 2009 by stonebro
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CyPhErIoN: hey m8 , i'm a design engineer so i could help you with this. what part do you want to elaborate on? i don't see d at first glanse.
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lowyhong: Sorry I meant delta instead of 'd'.

now that helps but seems thatstonebro nailed it already :)
Ah thanks both of you. Incidentally, I thought the page should have automatically moved down to Problem 4, but from the looks of it, you guys must have been searching for the question.
I also have a few more questions to raise.
1) To solve Question 4, is it absolutely necessary to assume an arbitrary value for δ?
2) Generally for precise limit questions, is it true that δ will usually be written as a multiple/fraction of ε, if the limit can be proven? Is it like working along a tangential line or something?
3) I understand δ = min{1,ε/3} is to find the smallest value of the set, but I don't understand why "this guarantees that both assumptions made about δ in the course of this proof are taken into account simultaneously." How does this actually guarantee that?
4) Why do we work backwards from ε instead of from δ? In other words, why do we determine δ from ε, but not the other way round?
5) I am looking at this site. The explanations are very clear cut, but what I don't understand is the verification part. For example in Example 2, why do we need to verify δ = ε/5 when we have already determined earlier on that |x-2| < ε/5? What is the point, because to me it looks like an extra unnecessary step - I can't help but think it's like saying "2 + 2 is 4, now let's verify that by adding the numbers one by one"
Post edited August 27, 2009 by lowyhong
Good lord, calculus? My brain hurts already.
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michaelleung: Good lord, calculus? My brain hurts already.

You don't know the other half of it. I'm not even a Math major. Despite some slight interest in Maths, therein lies my forte, in essay writing.
to write the essay you first have to understand the matter my friend :)
anyway i'll try to get back to you for this one , i have a meeting in 5mins
I've been out of touch with calculus/math for a couple decades, but I believe the verification is as simple as the fact that you're actually making a guess at the beginning in order to prove the limit. You can't prove anything by guessing, so you have to verify the guess in order to make the proof valid. Redundant, yes, but without verification your 'proof' remains only a guess.
It's like saying "2 + some number equals 4. I'm going to guess that number is 2." Then you have to prove the guess is correct.
Post edited August 27, 2009 by Coelocanth
I think the correct answer is:
No, nobody is good with calculus.
maybe ask your teacher/fellow student.
While I can answer each of 5 questions, It will be much faster and better to understand if you do it one on one and get direct feedback. So that if you don't understand a part you can directly ask for clarification.
Alright after a lot of talking to myself and asking redundant and useful questions, I think I've figured out the basics already. Thanks everybody! Reps given to all participants
Qbix: I'm planning to arrange for consultation sessions with my lecturer. But I'm lagging so far behind, I figured it's not going to hurt if I do some homework at home first ;)
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Coelocanth: I've been out of touch with calculus/math for a couple decades, but I believe the verification is as simple as the fact that you're actually making a guess at the beginning in order to prove the limit. You can't prove anything by guessing, so you have to verify the guess in order to make the proof valid. Redundant, yes, but without verification your 'proof' remains only a guess.
It's like saying "2 + some number equals 4. I'm going to guess that number is 2." Then you have to prove the guess is correct.

But if I can make δ the subject of ε, how is that guessing? Like, let's say I factorise my |f(x) - L| until I can choose δ = 1/2ε or something. Since I already have worked out that δ can = 1/2ε, why do I still need to verify one more time?
Post edited August 27, 2009 by lowyhong
i think that you don't need to really
it actually depends on the questions like when dealing with determinants.
You first have to make an assumption and then you have got to prove it.
once proven you can use this in other equasions (forgive the terminology since i'm dutch)
if you already worked out that δ = 1/2ε you don't need to proof it again.
as coelocanth said : 2 + ? = 4 now lets guess and proof.