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Proof that cos x diverges

  1. Aug 5, 2011 #1
    1. The problem statement, all variables and given/known data
    Recall that a function G(x) has the limit L as x tends to 1, written
    lim as x -> infinity
    G(x) = L,
    if for any epsilon > 0, there exists M > 0 so that if x > M , then
    |G(x) L| < epsilon.
    This means that the limit of G(x) as x tends to 1 does not exist if for
    any L and positive M, there exists epsilon > 0 so that for some x > M ,
    |G(x) L| >= epsilon.
    Using this definition, prove that

    integral sinx dx from 2 pi to infinity diverges.


    2. Relevant equations



    3. The attempt at a solution

    currently i just have no idea how to start this question up.
    knowing that the integral is equals to -cos x + C, while cosine value is bouncing between -1 and 1, how do i start the proof using that definition
     
  2. jcsd
  3. Aug 5, 2011 #2
    I think I just saw this question on Yahoo Answers lol
     
  4. Aug 5, 2011 #3
    Use the fact that certain multiples of x will give -1 and others will give 1 for x>M, thus you can't pick an arbitrary epsilon>0. That is, you can't get as close as you like to a limit L for all values x>M.
     
  5. Aug 5, 2011 #4

    HallsofIvy

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    Arcana Noir is referring to the fact that [itex]\lim_{x\to a} f(x)= L[/itex] if and only if [itex]\lim_{n\to\infty}f(x_n)= L[/itex] for every sequence [itex]\{a_n\}[/itex] such that [itex]\lim_{n\to\infty} a_n= a[/itex].

    Look at [itex]a_n= n\pi[/itex] and [itex]a_n= (2n+1)\pi/2[/itex].
     
  6. Aug 5, 2011 #5
    Thank you for clarifying that, HallsofIvy, I only just learned this sort of thing this week, and only in terms of a sequence. I have the ideas but not the technical form yet.
    The phrase "as close as you like to L" I picked up back in calc I, and have always preferred it conceptually to "an arbitrary epsilon >0".
     
    Last edited: Aug 5, 2011
  7. Aug 5, 2011 #6
    I think also we need to be a bit more careful with our definition of a limit and its negation.

    To say that a function G does not have ANY limit L as x tends towards infinity means that (1) for all L (2) there exists an [itex]\epsilon>0[/itex] such that (3) for all [itex]M>0[/itex] (4) there exists an x in the domain so that

    [tex] x>M[/tex] and [tex] |G(x)-L|\geq \epsilon.[/tex]

    Notice the sequence of quantifiers: (1) for all, (2) there exists, (3) for all, (4) there exists. This sequence is very important and if you change it the definition falls apart. Remember that the negation of [itex] \forall[/itex] is [itex] \exists[/itex] and the negation of [itex] \exists[/itex] is [itex] \forall[/itex].
     
  8. Aug 5, 2011 #7
    The integral is -cos x evaluated from 2π to ∞. In other words, it's:

    -cos(2π) + lim cos(r) as r approaches ∞.
    =-1+cos(∞)

    But cos(∞) is equal to the limit of cos(1/x) as x approaches 0. Make a power series for cos(x) and replace all the terms with 1/x and you'll see clearly that it diverges. Then you can apply your limit definition on the power series and show that it's impossible.
     
  9. Aug 5, 2011 #8

    Redbelly98

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    Moderator's note:

    Now that the OP has received several hints, please give bazingga123 a chance to work on the problem and reply before offering further help.
     
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