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Math Help - Proving a limit of at infinity based on certain given stipulations?

  1. #1
    Member mfetch22's Avatar
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    Proving a limit of at infinity based on certain given stipulations?

    Heres the question that I think I have figured out. I was hoping somebody could look over my proof and tell me weather it is valid. Anyway, the question is:



    LIMIT PROBLEM

    If \;\; 0 < \epsilon_1 < f(x) \;\; for all \; x,

    and

    \lim_{x \to a} g(x) = 0

    then prove that:

    \lim_{x \to a} \frac{f(x)}{|g(x)|} = \infty

    Okay, so heres my attempt at a proof. Please point out any errors you may see or find:



    MY ATTEMPT AT PROVING THE LIMIT

    By the definition of a limit, we know that:

    \lim_{x \to a} g(x) = 0

    Means that for every:
    \epsilon_2 > 0,theres is a \delta_1 > 0 such that:

    0 < |x - a| < \delta_1

    implies:

    |g(x) - 0| = |g(x)| < \epsilon_2

    The above inequality implies that:

    \frac{1}{\epsilon_2} < \frac{1}{|g(x)|}

    Also, since:

    0 < \epsilon_1 < f(x)
    Then it follows that (by multiplying \epsilon_1 to both sides of the g(x) inequality):

    \frac{\epsilon_1}{\epsilon_2} < \frac{f(x)}{|g(x)|}

    [I don't know if the follwing is neccesary for this proof, but better safe then sorry]

    (We need not worry about \frac{f(x)}{|g(x)|} being negitive; since 0 \leq |a| for all a, then the denominator is always positive. And since, by the stipulations made by the problem, 0 < \epsilon_1 < f(x) for all x obviously means that 0 < f(x) for all x; then it folllows that both the denominator and numerator are positive, therefore 0 < \frac{f(x)}{|g(x)|} for all x)

    To prove that the limit approaches infinity, we must show that:

    0 < |x - a| < \delta_2

    Implies that, for any N we have:

    N < \frac{f(x)}{|g(x)|}.

    But we are already there! (atleast I think, as long as there are no errors) From the given limit of g(x), we were able to show that :

    \frac{\epsilon_1}{\epsilon_2} < \frac{f(x)}{|g(x)|}

    So letting:

    N = \frac{\epsilon_1}{\epsilon_2}

    and

    \delta_1 = \delta_2

    and rememebering the stipulations made by the problem of the value of f(x), then the train of implications goes like this:

    0 < |x - a| < \delta_2

    \Rightarrow \;\;\; 0 < |x - a| < \delta_1

    \Rightarrow \;\;\; |g(x) - 0| = |g(x)| < \epsilon_2

    \Rightarrow \;\;\; \frac{1}{\epsilon_2} < \frac{1}{|g(x)|}

    Remeber that:

    0 < \epsilon_1 < f(x)

    So:

    [0 < \epsilon_1 < f(x)] \; \wedge \; [\frac{1}{\epsilon_2} < \frac{1}{|g(x)|} ]

    \Rightarrow \;\;\; \frac{\epsilon_1}{\epsilon_2} < \frac{f(x)}{|g(x)|}

    Which shows that:

    0 < |x-a| < \delta_2 \;\; \Rightarrow \;\; N < \frac{f(x)}{|g(x)|}

    For:

    N = \frac{\epsilon_1}{\epsilon_2}

    and

    \delta_1 = \delta_2

    FINNALY, this all means that:

    \lim_{x \to a} \frac{f(x)}{|g(x)|} = \infty

    (granted that the stipulations on f(x) are given)


    Now, are there any errors in this proof? And is it always neccesary to mathematically go over every detail in a proof? Like, is my proof to long? Is it okay to make simple assumptions that your thinking isn't neccesary? (Like stating or not stating in the proof that 0 \leq |a| for all a) Thanks in advance
    Last edited by mfetch22; August 20th 2010 at 11:55 AM.
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  2. #2
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    Opalg's Avatar
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    That argument is essentially correct. The only thing I would criticise is that you need to be clear about what is given and what is subsequently chosen. The quantity \epsilon_1 is given in the statement of the problem. The quantity N is then given as a "target" in the definition of the limit being infinity. The quantity that is then chosen by you during the course of the proof is \epsilon_2. So in the middle of the proof it would be better to write "Let \epsilon_2 = \epsilon_1/N" rather than letting N = \epsilon_1/\epsilon_2.
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  3. #3
    Member mfetch22's Avatar
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    Quote Originally Posted by Opalg View Post
    That argument is essentially correct. The only thing I would criticise is that you need to be clear about what is given and what is subsequently chosen. The quantity \epsilon_1 is given in the statement of the problem. The quantity N is then given as a "target" in the definition of the limit being infinity. The quantity that is then chosen by you during the course of the proof is \epsilon_2. So in the middle of the proof it would be better to write "Let \epsilon_2 = \epsilon_1/N" rather than letting N = \epsilon_1/\epsilon_2.
    Thank you for that clarification. But I'm curious about one detail. I trust your judgement extremely more then mine concerning these matters of mathematical proofs, since I'm quite new to learning real analysis; but I just wanted to make 100% sure of my understanding.

    You implied that it'd be more appropriate "to write 'Let \epsilon_2 = \epsilon_1 / N \; rather then N = \epsilon_1 / \epsilon_2". Remebering what was said in the textbook I've been using over the summer, I thought that (and this may just be my mistunderstanding of the limit chapter) but I thought that the definition of a limit, and the given information implies that:

    \epsilon_2 = \epsilon (\delta_1)

    Where \; \epsilon(\delta_1) \; is a function of \delta_1

    ?

    What I mean is, since the problem provided that:

    \lim_{x \to a} g(x) = 0

    Or in other words:

    0 < | x - a| < \delta_1 \; \; \Rightarrow \; \; |g(x)| < \epsilon_2

    And since we are given the fact that the left side of the statement above implies the right side, is it neccesarily valid to assume from this (assuming from the given information that is) that \epsilon_2 can be written as a function of \delta_1 ?

    If this is a valid assumption, then would it be correct or incorrect to still just define:

    N = \epsilon_1 / \epsilon_2

    Since the infromation given in the problem implies that \epsilon_2 already has a definition?

    Or, since a definition was not formmally given at any point in the proof (it was only implied to exist, if I was correct in my assumption), is it required by rigor to place within the proof a definition of \epsilon_2 ? Which, as you said would be:

    \epsilon_2 = \epsilon_1 / N

    ?

    Thank you for your help
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  4. #4
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by mfetch22 View Post
    I thought that the definition of a limit, and the given information implies that:

    \epsilon_2 = \epsilon (\delta_1)

    Where \; \epsilon(\delta_1) \; is a function of \delta_1

    ?
    Wrong way round! The definition of a limit says that given \epsilon>0 there exists \delta>0 such that (blah blah blah). So \delta is a function of \epsilon, not the other way round.

    In your problem, you want to show that if x is sufficiently close to a, then f(x)/|g(x)| can be made larger than some given N. That will be the case provided that |g(x)| can be made less than \epsilon_2 = \epsilon_1/N. And the definition of \lim_{x\to a}g(x)=0 says that this in turn can be achieved provided that |x-a|<\delta_2 for some \delta_2 depending on \epsilon_2.
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  5. #5
    Member mfetch22's Avatar
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    Quote Originally Posted by Opalg View Post
    Wrong way round! The definition of a limit says that given \epsilon>0 there exists \delta>0 such that (blah blah blah). So \delta is a function of \epsilon, not the other way round.

    In your problem, you want to show that if x is sufficiently close to a, then f(x)/|g(x)| can be made larger than some given N. That will be the case provided that |g(x)| can be made less than \epsilon_2 = \epsilon_1/N. And the definition of \lim_{x\to a}g(x)=0 says that this in turn can be achieved provided that |x-a|<\delta_2 for some \delta_2 depending on \epsilon_2.
    Ohhhhhhhhhh! I seee! Your completely correct! For some reason it seems to make much more sense now, in the way you worded it above. Thank you for your assistance
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