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Proving the existence of a limit?

  1. Feb 17, 2008 #1
    How do you prove the existence but not necessarily the value of a limit?
  2. jcsd
  3. Feb 17, 2008 #2
    Why not give an example of such a problem because I think your question is too general.
  4. Feb 17, 2008 #3
    Suppose you are given a limit. Show that a limit exists without needing to say what it is.

    It is a general question.
  5. Feb 17, 2008 #4
  6. Feb 17, 2008 #5
    Okay, good point. I'll be more specific. I was thinking of limits of sequences only. Or are all limits sequences in which case I haven't narrowed down anything?
  7. Feb 17, 2008 #6
    Well how about showing the limit of [tex]a_n-a_{n+1}=0[/tex] as n->oo
  8. Feb 17, 2008 #7


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    This doesn't imply that (a_n) will converge. [Counterexample: a_n = log(n).]
  9. Feb 17, 2008 #8


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    If you're going to be taking a limit of something, then it had better be a sequence (or something sequence-like). So, no, you haven't narrowed down anything. Do you mean sequences of numbers?

    In any case, there are a few existence results that can come in handy. One example is the monotone convergence theorem, which states that if a sequences of real numbers is monotone and bounded, then it must converge.
  10. Feb 17, 2008 #9
    What if a_n is bound both above and bellow?
  11. Feb 17, 2008 #10
    In which case it would be in a compact space. All cauchy sequences in a compact space converge.
  12. Feb 17, 2008 #11


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    Just because (a_n - a_{n+1}) -> 0 doesn't mean a_n is Cauchy. Again, a counterexample to this claim is a_n = log(n). We have log(n) - log(n+1) = log(n/(n+1)) -> log(1) = 0, but if this were Cauchy, then it would be convergent (since the reals are complete), and it clearly isn't.

    It turns out that being bounded isn't good enough either, although finding a counterexample was trickier. At any rate: try a_n = exp(i(1 + 1/2 + ... + 1/n)). This doesn't converge - it goes around the unit circle in the complex plane. On the other hand,
    [tex]a_n - a_{n+1} = \exp\left(i \left(1 + \frac{1}{2} + ... + \frac{1}{n}\right)\right)\left(1 - \exp\left(\frac{i}{n+1}\right)\right) \to 0.[/itex]
    Last edited: Feb 17, 2008
  13. Feb 17, 2008 #12
    Okay, how about [tex]a_n-a_{n-1} \le k \left( \frac{1}{n}-\frac{1}{n-1} \right) [/tex] for all n greater then M
    Last edited: Feb 17, 2008
  14. Feb 18, 2008 #13
    That's a pretty interesting example. Can you find a similar example on the reals?
  15. Feb 18, 2008 #14


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    What are k and M?

    Take the real part of it and see what happens.
  16. Feb 18, 2008 #15
    To be more specific, I had a sequence a_m/b_m where a_m and b_m in the limit is 0.

    However they are not functions so can't use that rule which I can't spell, l'hospil's rule?
    Last edited: Feb 18, 2008
  17. Feb 18, 2008 #16
    k an M are arbitrary chosen to make the inequality work for a given sequence.
  18. Feb 18, 2008 #17


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    Why didn't you say that when you were first asked to be specific?

    If all you know is that [itex]a_n[/itex] and [itex]b_n[/itex] go to 0, you can't say anything about whether [itex]a_n/b_n[/itex] converges or diverges. For example, if [itex]a_n= 1/n^2[/itex] and [itex]b_n= 1/n[/itex], it is obvious that both [itex]a_n[/itex] and [itex]b_n[/itex] converge to 0. And [itex]a_n/b_n= (1/n^2)(n/1)= 1/n[/itex] converges to 0.

    But if [itex]b_n= 1/n^2[/itex] and [itex]a_n= 1/n[/itex], it is still obvious that both [itex]a_n[/itex] and [itex]b_n[/itex] converge to 0 but now [itex]a_n/b_n= (1/n)(n^2/1)= n[/itex] does not converge.
  19. Feb 20, 2008 #18


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    Looking upon the 1st post: if left hand and right hand limits are equal, then the limit exists.
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