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Find the limit of sequence (trig term)

  1. Feb 8, 2013 #1
    1. The problem statement, all variables and given/known data

    find the limit of this sequence

    1) An = [itex]\frac{\sqrt{n} sin (n! e^{n})}{n+1}[/itex]

    2) An = [itex] 4n^{3} sin\frac{1}{n^{3}} [/itex]

    2. Relevant equations



    3. The attempt at a solution
    (1)
    |sin n! e^n)| ≤1

    |An| ≤ [itex]\frac{ \sqrt{n} }{n+1}[/itex]

    |An| ≤ [itex]\frac{ \sqrt{n}/n }{n/n+1/n}[/itex] = 0

    lim[itex] _{n\rightarrow \infty}[/itex] An = 0 √

    ==========================

    (2)

    [itex]|sin \frac{1}{n^{3}}| [/itex] ≤ 1
    An ≤ [itex] 4 n^{3} [/itex]
    lim[itex] _{n\rightarrow \infty}[/itex] An = ∞ X

    The second question my answer is wrong
    I should get 4. and the answer shows that I have to use L' Hospital's
    My question is why I cannot apply the method from the first question to the second question.
    It just confused me I am not sure but my solution on the first question anyway but I got the right answer please comment on that too please !!!

    thank you for your help
     
  2. jcsd
  3. Feb 8, 2013 #2

    SammyS

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    As n→∞, 1/n3 → 0 so sin(1/n3) → 0.

    So that, 4n3 sin(1/n3) is indeterminate of the form ∞∙0 .

    Even if have a sequence such as [itex]\displaystyle 4n^3\sin(n)[/itex] your method will give an incorrect result because sin(n) oscillates from near -1 to near +1 .
     
    Last edited: Feb 8, 2013
  4. Feb 8, 2013 #3
    For the second question, replace n with 1/h where h->0. Then use L'Hopital's Rule.
     
  5. Feb 8, 2013 #4
    What I don't understand is why it works on the first question but the second question. still don't get it. thanks for the comments
     
  6. Feb 8, 2013 #5
    why we can set h->0 I have seen that method but I don't get it why n-> infinity can change to h->0. sorry if I asked such a stupid question. thanks for the comments
     
  7. Feb 8, 2013 #6

    haruspex

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    The question has 1/n3, not 1/n2.
    izen, lim n→∞ n3 sin(1/n3) must be the same as lim n→∞ n sin(1/n), right? Does that help?
    It's just a change of variable. Define h = 1/n, then n sin(1/n) = sin(h)/h, and the limit of n sin(1/n) as n→∞ must be the same as lim sin(h)/h as h→0. Having converted it to a problem with a finite limit point you can use l'Hopital's rule.
     
  8. Feb 8, 2013 #7
    I need time to think about your say. Thank you so much haruspex :)
     
  9. Feb 8, 2013 #8
    so it means lim n→∞ 1/n = 0 so if we define h=1/n. the limit of 'h' needs to be equal to 1/n by setting lim h→0 h = 0. Am i understand it right ?
     
  10. Feb 8, 2013 #9

    SammyS

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    Consider a sequence, {An}, such that the each individual term is given by,
    An = Bn∙Cn
    where {Bn} is bounded above and below, but does not converge.

    If {Cn} converges to zero, then so does {An}.

    Otherwise, {An} does not converge -- no matter whether {Cn} converges or not.



    Your first problem is a case of the above.


    Your second problem does not nearly fit the above requirements.

    Yes, the sequence {sin(1/n3)} is bounded, but more than that, this sequence converges to zero.* The sequence {4n3} → +∞ as n → ∞ . So, the term by term product of these sequences may or may not converge based upon other totally different criteria.
    * The sequence {1/n3} converges to zero, and sin(0) = 0 .
     
    Last edited: Feb 8, 2013
  11. Feb 8, 2013 #10
    In your first question you show that your sequence lies beetween two sequence that both goes to zero 0≤An≤0 this implies An → 0
    It's important that you show that it lies BEETWEEN two sequences.

    In your secondo question you show that your question is less than infinity, that does not make it go to infinity though. An could go 2, wich is less than infinity, or -10 which is less than infinity, or -∞ which is less than infinity.
     
  12. Feb 8, 2013 #11

    haruspex

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    Yes.
     
  13. Feb 8, 2013 #12
    i get it now
    thank you u guys all
     
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