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Math Help - limits with e

  1. #1
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    limits with e

    find the limit of a_n = \frac{e^n + e^{-n}}{e^{2n} - 1} an n approaches infinity

    L'Hop's rule doesnt work and dividing the terms by e^2n doesnt work either. I do not know what process should be used.
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  2. #2
    MHF Contributor chisigma's Avatar
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    You have only to consider that is ...

    a_{n} = \frac { e^{n} + e^{-n}}{ e^{2n}-1} = e^{-n}\cdot \frac{ e^{n} + e^{-n}}{ e^{n}-e^{-n}} (1)

    ... and then find the limit when n tends to infinity of the two term of (1)... not very difficult...

    Kind regards

    \chi \sigma
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  3. #3
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     e^{-n}\cdot \frac{ e^{n} + e^{-n}}{ e^{n}-e^{-n}}
    I don't understand how you got that part.
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  4. #4
    MHF Contributor chisigma's Avatar
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    a_{n} = \frac { e^{n} + e^{-n}}{ e^{2n}-1} = \frac{ e^{n} + e^{-n}}{e^{n}\cdot (e^{n}-e^{-n})}


    ... all right?...


    Kind regards


    \chi \sigma
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  5. #5
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    Another way to do this is to divide both numerator and denominator by e^{2n}:
    \frac{e^{-n}+ e^{-3n}}{1- e^{-2n}}
    Now what happens to all negative powers as n goes to infinity?
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  6. #6
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    @HallsofIvy
    I did that but i guess my algebra was off. Your process works, thanks.
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