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Epslion Proof for limit of a sequence

  1. Feb 9, 2008 #1
    1. The problem statement, all variables and given/known data

    Using only the definition of a limit of a sequence prove that lim n->infinity tanh(n)=1


    2. Relevant equations



    3. The attempt at a solution

    My attempt at the solution is as follows.

    If 1 is the limit of the sequence then for every [tex]\epsilon[/tex]>0, there exists an number such that n>N for every n, such that we have

    |tanh(n)-1|<[tex]\epsilon[/tex]
    apply the appropriate hyperbolic identity I re-write this as.

    |e^2n-1/(e^2n+1) -1 |< [tex]\epsilon[/tex]
    as tanh(n)< 1 for every sufficiently large n

    we have 1-(e^2n-1/(e^2n+1)) < [tex]\epsilon[/tex]

    After this im stumped, our textbook is very poor so are the notes.

    Any help is welcome thanks in advanced.
     
  2. jcsd
  3. Feb 9, 2008 #2
    How about factoring out e^2n, and then applying limit, and if you are missing lim n-->inf sign there?
     
  4. Feb 9, 2008 #3
    Re: factoring out

    If i could evaluate it as a limit I would but its asking to prove it using the precise definition of a limit of a sequence, which means that is not an option
     
  5. Feb 9, 2008 #4

    lurflurf

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    lim n->infinity tanh(n)=1
    we desire to show
    for any eps>0
    there exist (a natural number) N(eps)
    such that whenever (a natural number) n>N(eps)
    |1-tanh(n)|<eps
    hint
    show |1-tanh(n)|<2exp(-2n)
    Then the original question is equivalant to showing
    lim n->infinity exp(-n)=0
     
  6. Feb 9, 2008 #5

    lurflurf

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    you we so close
    you had
    (1-(e^2n-1/(e^2n+1))
    ((e^2n+1)/(e^2n+1)-(e^2n-1)/(e^2n+1))
    ((e^2n+1)-(e^2n-1))/(e^2n+1)
    2/(e^2n+1)
    2/(e^2n+1)<2exp(-n)
    thus the hint
    it should be easy to finish
     
  7. Feb 9, 2008 #6
    Try this

    [tex]\left|\frac{e^{2n}-1}{e^{2n}+1}-1\right|=\left|\frac{-2}{e^{2n}+1}\right|=\frac{2}{e^{2n}+1}<\frac{2}{e^{2n}}=2\,e^{-2n}<\epsilon[/tex]

    and choose
    [tex]N=max\left\{0,-\frac{1}{2}\,\ln\frac{\epsilon}{2}\right\}[/tex]
     
  8. Feb 9, 2008 #7
    Oupps! lurflurf was faster! :smile:
     
  9. Feb 9, 2008 #8
    Thanks guys working it through now.
     
  10. Feb 9, 2008 #9
    I am currently in calc II. I am curious where you got what we choose N= , I can finish it with the substitution you gave me. I am just curious how one derives it.
     
  11. Feb 9, 2008 #10
    And I got the problem thank you, However I am still curious about the derivation of N
     
  12. Feb 9, 2008 #11
    There is no general method to choose [itex]N[/itex]. One starts with the definition [itex]|a_n-l|<\epsilon[/itex] and tries to find [itex]N[/itex]. In the problem at hand, since we don't know if [itex]-\frac{1}{2}\,\ln(\epsilon/2)[/itex] is positive we have to write [itex]N=max\left\{0,-\frac{1}{2}\,\ln(\epsilon/2)\right\}[/itex]
     
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