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    Hi, I have the following question:

    Name:  Capture.PNG
Views: 144
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    I did the proof as dictated by a book, and have the following:

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    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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    (Original post by raees)
    Hi, I have the following question:

    Name:  Capture.PNG
Views: 144
Size:  1.4 KB

    I did the proof as dictated by a book, and have the following:

    Name:  Capture.PNG
Views: 151
Size:  6.8 KB

    Name:  Capture.PNG
Views: 108
Size:  7.8 KB

    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
    The mistake I noticed was that the final line in the 1st image should use the fact that

    \displaystyle -\epsilon - 4 < -x < \epsilon - 4  \Rightarrow \epsilon + 4 > x > 4 -\epsilon

    Also, it's always better to use quantifiers, at undergraduate level anyway.
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    (Original post by Indeterminate)
    The mistake I noticed was that the final line in the 1st image should use the fact that

    \displaystyle -\epsilon - 4 < -x < \epsilon - 4  \Rightarrow \epsilon + 4 > x > 4 -\epsilon

    Also, it's always better to use quantifiers, at undergraduate level anyway.
    quantifiers?
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    (Original post by raees)
    quantifiers?
    \forall

    means "for all"

    \exists

    means "there exists"

    As the name suggests, they relate to quantity.
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    oh! ok, thank you for the help
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    (Original post by Indeterminate)
    \forall

    means "for all"

    \exists

    means "there exists"

    As the name suggests, they relate to quantity.
    Do not listen to this guy. Words are vastly preferable to symbols. Symbols are just abbreviations; sometimes they are useful and sometimes they aren't.

    For instance

    e^{i\pi}

    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 \epsilon > 0 there exists a \delta > 0 such that for all x with 0 < |x − c | < \delta, we have |f(x) − L| < \epsilon

    is vastly preferable to

    \forall \epsilon > 0 \exists\delta > 0\colon \forall x, 0 < |x  -  c | < \delta \Rightarrow|f(x)  - L| < \epsilon

    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.
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    (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
    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 \epsilon > 0 be given and define \delta := \epsilon.

    Then 0 < |x-4| < \delta is equivalent to 0 < |4-x| < \delta (by definition of the modulus) which is further equivalent to  0 < |4-x| < \epsilon (since \epsilon = \delta)

    The final inequality may be written as  0 < |f(x)-L| <\epsilon

    where f(x) = 9-x and L=5.

    Therefore, by definition, the limit of f(x) as x \rightarrow 4 is 5.


    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).
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    (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 \epsilon > 0 be given and define \delta := \epsilon.

    Then 0 < |x-4| < \delta is equivalent to 0 < |4-x| < \delta (by definition of the modulus) which is further equivalent to  0 < |4-x| < \epsilon (since \epsilon = \delta)

    The final inequality may be written as  0 < |f(x)-L| <\epsilon

    where f(x) = 9-x and L=5.

    Therefore, by definition, the limit of f(x) as x \rightarrow 4 is 5.


    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).
    Now I'm a bit confused :/

    I looked at the next question it asks for me to prove the following:

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    So I do as you say and get the following:

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    but then where do I go from here?

    p.s. sorry for late reply, didn't get a notification that someone else had replied.
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    (Original post by raees)

    but then where do I go from here?
    I don't think you understood what I was saying. The game with these proofs is that if I give you an \epsilon >0, you give me a \delta such that if x is within \pm \delta of a point c, then f(x) is within \pm \epsilon of L. If you can do that for any \epsilon >0 I give you, then you have shown that \mathrm{lim}_{x \rightarrow c} f(x) = L.

    Generally, the way you write a proof covering any choice of \epsilon is to write \delta as a function of \epsilon.

    The sole aim of the game is working out how to define \delta in a way that works. The method that you used to work out \delta was fine and basically involves going backwards to ensure that the \delta 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 \delta 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 \delta and then write the proof. Here, I don't think \delta:=\epsilon will work. You need to go back and use some method to define a \delta that will work.

    Hint: To simplify things, put y:=x-5. This means, you want to define \delta such that

    0<\mid y- 4\mid < \delta \Rightarrow 0 < \mid \sqrt{y} - 2 \mid < \epsilon

    Further hint: Use the fact that

     0 < \mid \sqrt{y} - 2 \mid < \epsilon
    \Leftrightarrow 0 < \mid \sqrt{y} - 2 \mid \frac{\mid \sqrt{y} + 2 \mid}{\mid \sqrt{y} + 2 \mid} < \epsilon
    \Leftrightarrow 0 < \mid \sqrt{y} - 2 \mid \mid \sqrt{y} + 2 \mid < \epsilon \mid \sqrt{y} + 2 \mid
    \Leftrightarrow 0 < \mid y - 4 \mid < \epsilon \mid \sqrt{y} + 2 \mid

    Further further hint:

    If \delta<1 then

    \mid y - 4 \mid < \delta \Rightarrow 3 < y < 5 \Rightarrow \sqrt{3} + 2 < \sqrt{y} + 2 < \sqrt{5} +2

    Can you put these together and work out how to define suitable delta?
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    (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 \epsilon >0, you give me a \delta such that if x is within \pm \delta of a point c, then f(x) is within \pm \epsilon of L.
    How do I know what the value of x will be?
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    (Original post by raees)
    How do I know what the value of x will be?
    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 \pm \epsilon by taking any value of x within \pm \delta of the point c. 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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    (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 \pm \epsilon by taking any value of x within \pm \delta of the point c. 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.
    ok, thanks for the help anyways
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    (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 \pm \epsilon by taking any value of x within \pm \delta of the point c. 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.
    Ok so I did some more studying and have run into a problem with a question.

    This is the question:

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    and I did the following first(rough work), to determine a value for delta:

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    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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    (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?
    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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    (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.
    ook... how about this -->

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    (Original post by raees)
    ook... how about this -->
    You seem to be going along the right lines in your workings but your 'proof' makes no logical sense and is incomplete. You need to actually write a sentence or two, starting with a premise and arriving at a conclusion and try to explain what is going on where necessary.

    Another problem is also that your choice of delta won't work here. Your workings relied on \delta = \mathrm{min}\{1,\epsilon(\sqrt{5  } +2)\} but then you unexplicably choose to define \delta=\epsilon(\sqrt{5} +2)... You also needed to have used the lower of the bounds on x rather than the higher to make it work so would need \delta = \mathrm{min}\{1,\epsilon(\sqrt{3  } +2)\}

    Think about it, the proof has to work for any positive \epsilon. If \epsilon is larger than \frac{1}{\sqrt{3} +2} then your bound on x doesn't apply anymore; you assumed that \delta was less than 1 for that.

    To write a proof that a limit exists and has a certain value like this you have to start by fixing an \epsilon &gt; 0, defining a \delta (depending on epsilon) and then assuming that, in this case, 0 &lt; |x-9| &lt; \delta and then deriving from that the conclusion |\sqrt{x-5}-2|&lt;\epsilon 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 \epsilon &gt; 0 and set \delta:=\mathrm{min}\{1,\epsilon  (\sqrt{3} +2)\}.

    Then 0&lt;|x-9|&lt;\delta \Rightarrow \cdots \Rightarrow |\sqrt{x-5}-2|&lt; \frac{\epsilon(\sqrt{3} +2)}{\sqrt{x-5}+2} since \delta \leq \epsilon(\sqrt{3} +2).

    On the other hand, 0&lt;|x-9|&lt;\delta \Rightarrow \cdots \Rightarrow \sqrt{x-5}+2 &gt; \sqrt{3} +2 since \delta \leq 1.

    Therefore, \frac{\sqrt{3} +2}{\sqrt{x-5}+2} &lt;\frac{\sqrt{3} +2}{\sqrt{3}+2}=1 and so \frac{\epsilon(\sqrt{3} +2)}{\sqrt{x-5}+2}&lt;\epsilon. Together with the above, we get that
    0&lt;|x-9|&lt;\delta \Rightarrow |\sqrt{x-5}-2|&lt; \epsilon as required.
 
 
 
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