Hi, I have the following question:
I did the proof as dictated by a book, and have the following:
Could someone please check if it's ok, because there is no solutions in the book. Also I'll be uploading other exercises after this so just please check if it's ok

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 09032013 16:47

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 09032013 17:12
(Original post by raees)
Hi, I have the following question:
I did the proof as dictated by a book, and have the following:
Could someone please check if it's ok, because there is no solutions in the book. Also I'll be uploading other exercises after this so just please check if it's ok
Also, it's always better to use quantifiers, at undergraduate level anyway.Last edited by Indeterminate; 09032013 at 17:31. 
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 09032013 18:14
(Original post by Indeterminate)
The mistake I noticed was that the final line in the 1st image should use the fact that
Also, it's always better to use quantifiers, at undergraduate level anyway. 
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 09032013 18:24
(Original post by raees)
quantifiers?
means "for all"
means "there exists"
As the name suggests, they relate to quantity. 
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 09032013 20:02
oh! ok, thank you for the help

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 09032013 23:49
(Original post by Indeterminate)
means "for all"
means "there exists"
As the name suggests, they relate to quantity.
For instance
is much better than
"The exponential function evaluated at the square root of minus one times the area of the unit circle"
but on the other hand
for all there exists a such that for all with , we have
is vastly preferable to
since it is eminently more readable.
Save the overuse of symbols for notes and formal logic and use words in your own writing. This is how maths is written  take a look at any book.Last edited by Mark85; 19032013 at 15:02. Reason: latex 
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 10032013 00:03
(Original post by raees)
Could someone please check if it's ok, because there is no solutions in the book. Also I'll be uploading other exercises after this so just please check if it's ok
When you are writing an epsilon delta type proof, you should start off by fixing epsilon and defining delta and showing that it works.
What the book has told you is the way in general to find a delta that works. What you have written out is your workings out/scrap notes that you would calculate BEFORE writing out an actual answer. When you write out an answer, you do it in the way easiest to read and follow, not in the order following the thought process needed to get to it.
In this case, you have overcomplicated something extremely simple. For example, here is my proof.
Let be given and define .
Then is equivalent to (by definition of the modulus) which is further equivalent to (since )
The final inequality may be written as
where and .
Therefore, by definition, the limit of as is .
The point was that both inequalities were the same with epsilon replaced with delta so choosing delta equal to epsilon makes the RHS equivalent to the LHS (which is even stronger than what you need).Last edited by Mark85; 10032013 at 00:08. 
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 13032013 16:13
(Original post by Mark85)
It is kind of ok (a couple of slip ups) but here is the main problem. The order and the way you have written it out is horrendous.
When you are writing an epsilon delta type proof, you should start off by fixing epsilon and defining delta and showing that it works.
What the book has told you is the way in general to find a delta that works. What you have written out is your workings out/scrap notes that you would calculate BEFORE writing out an actual answer. When you write out an answer, you do it in the way easiest to read and follow, not in the order following the thought process needed to get to it.
In this case, you have overcomplicated something extremely simple. For example, here is my proof.
Let be given and define .
Then is equivalent to (by definition of the modulus) which is further equivalent to (since )
The final inequality may be written as
where and .
Therefore, by definition, the limit of as is .
The point was that both inequalities were the same with epsilon replaced with delta so choosing delta equal to epsilon makes the RHS equivalent to the LHS (which is even stronger than what you need).
I looked at the next question it asks for me to prove the following:
So I do as you say and get the following:
but then where do I go from here?
p.s. sorry for late reply, didn't get a notification that someone else had replied.Last edited by raees; 13032013 at 16:14. 
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 13032013 18:31
Generally, the way you write a proof covering any choice of is to write as a function of .
The sole aim of the game is working out how to define in a way that works. The method that you used to work out was fine and basically involves going backwards to ensure that the does what you want but my point was that you can't write down all of your workings out as an answer in the order you worked it out  it makes for confused and horrible reading. Once you have worked out how to define you just do that at the start of the proof and then the rest of the proof is demonstrating that it works.
You can't generally start writing the proof straight away  you do some working out, figure out how to define and then write the proof. Here, I don't think will work. You need to go back and use some method to define a that will work.
Hint: To simplify things, put . This means, you want to define such that
Further hint: Use the fact that
Further further hint:
If then
Can you put these together and work out how to define suitable delta?Last edited by Mark85; 13032013 at 18:35. 
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 13032013 19:21
(Original post by Mark85)
I don't think you understood what I was saying. The game with these proofs is that if I give you an , you give me a such that if is within of a point , then is within of . 
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 13032013 21:10
(Original post by raees)
How do I know what the value of x will be?
You are showing that you can get the value of a function as close as you like to the limit (lying at a distance of by taking any value of within of the point . Try to think of the picture.
You really have to take the time to read, learn and inwardly digest the definition of a limit before you start being able to work out and write a proof that a limit exists. Otherwise, you will get very confused by what the book (or me or whoever) is actually trying to do.Last edited by Mark85; 13032013 at 21:13. 
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 13032013 21:16
(Original post by Mark85)
You are considering a limit at a particular point inside the domain at your function. The values of x are the ones in some neighbourhood of that point. Go back and read the definition of a limit and try to understand what it is...
You are showing that you can get the value of a function as close as you like to the limit (lying at a distance of by taking any value of within of the point . Try to think of the picture.
You really have to take the time to read, learn and inwardly digest the definition of a limit before you start being able to work out and write a proof that a limit exists. Otherwise, you will get very confused by what the book (or me or whoever) is actually trying to do. 
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 16032013 17:40
(Original post by Mark85)
You are considering a limit at a particular point inside the domain at your function. The values of x are the ones in some neighbourhood of that point. Go back and read the definition of a limit and try to understand what it is...
You are showing that you can get the value of a function as close as you like to the limit (lying at a distance of by taking any value of within of the point . Try to think of the picture.
You really have to take the time to read, learn and inwardly digest the definition of a limit before you start being able to work out and write a proof that a limit exists. Otherwise, you will get very confused by what the book (or me or whoever) is actually trying to do.
This is the question:
and I did the following first(rough work), to determine a value for delta:
I end up with two values.. I tried setting one, but I don't end up with
f(x)L<epsilon , how do I set a value? 
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 16032013 18:15
someone?

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 17032013 15:14
(Original post by raees)
I end up with two values.. I tried setting one, but I don't end up with
f(x)L<epsilon , how do I set a value?
But if you go back to my previous post, you will see that I found that if I assume that delta is less than one, I could then show that if delta was a particular multiple of epsilon, it would work. In that situation therefore, I would define delta to be the minimum of 1 and the particular multiple of epsilon that I determined so that both conditions would be satisfied. Without looking at what you have done, try taking delta to be the minimum of the two values and see if you can write out a proper proof (in here if you like) and we can help check it or otherwise, you will get to a point where it won't work and you can hopefully see the problem.Last edited by Mark85; 17032013 at 15:17. 
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 18032013 18:48
(Original post by Mark85)
To be honest, I can't really understand what you wrote... for instance I don't know why you would make delta be a function of x...
But if you go back to my previous post, you will see that I found that if I assume that delta is less than one, I could then show that if delta was a particular multiple of epsilon, it would work. In that situation therefore, I would define delta to be the minimum of 1 and the particular multiple of epsilon that I determined so that both conditions would be satisfied. Without looking at what you have done, try taking delta to be the minimum of the two values and see if you can write out a proper proof (in here if you like) and we can help check it or otherwise, you will get to a point where it won't work and you can hopefully see the problem.

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 19032013 13:34
Another problem is also that your choice of delta won't work here. Your workings relied on but then you unexplicably choose to define ... You also needed to have used the lower of the bounds on rather than the higher to make it work so would need
Think about it, the proof has to work for any positive . If is larger than then your bound on doesn't apply anymore; you assumed that was less than for that.
To write a proof that a limit exists and has a certain value like this you have to start by fixing an , defining a (depending on epsilon) and then assuming that, in this case, and then deriving from that the conclusion if necessary going backwards through your working to justify each step. Thus the form of the proof should look like this (and the dots between implications represent steps which must be filled in):
Fix and set .
Then since .
On the other hand, since .
Therefore, and so . Together with the above, we get that
as required.Last edited by Mark85; 19032013 at 13:35.
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