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Math Help - Sequences - limits

  1. #1
    Junior Member symstar's Avatar
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    Sequences - limits

    This is the correct answer, but I would appreciate it if someone would verify that the work is correct, particularly, is it okay to multiply through by natural log like I did?

    a_n = \frac{e^n+e^{-n}}{e^{2n}-1}

    \lim_{n \to \infty}a_n = \lim_{n \to \infty}\frac{e^n+e^{-n}}{e^{2n}-1} * \frac{\ln}{\ln}

    L=\lim_{n \to \infty}\frac{n-n}{2n}

    L=0
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by symstar View Post
    This is the correct answer, but I would appreciate it if someone would verify that the work is correct, particularly, is it okay to multiply through by natural log like I did?

    a_n = \frac{e^n+e^{-n}}{e^{2n}-1}

    \lim_{n \to \infty}a_n = \lim_{n \to \infty}\frac{e^n+e^{-n}}{e^{2n}-1} * \frac{\ln}{\ln}

    L=\lim_{n \to \infty}\frac{n-n}{2n}

    L=0
    no, it is not correct..
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  3. #3
    Junior Member symstar's Avatar
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    Do you have another suggestion? That was kinda my "last ditch" effort.
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  4. #4
    MHF Contributor kalagota's Avatar
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    a_n = \frac{e^n + e^{-n}}{e^{2n}-1} = \frac{e^n + e^{-n}}{e^{n}(e^n-e^{-n})} = \frac{\coth n}{e^n}
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  5. #5
    Junior Member symstar's Avatar
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    Thanks! Is it possible to do this without using hyperbolic functions though?
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  6. #6
    MHF Contributor kalagota's Avatar
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    personally, i don't know.. maybe others know it.. hehe
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  7. #7
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    Quote Originally Posted by symstar View Post
    This is the correct answer,
    a_n = \frac{e^n+e^{-n}}{e^{2n}-1}
    L=0 is correct but your method is not correct.
    \frac{{e^n  + e^{ - n} }}<br />
{{e^{2n}  - 1}} = \frac{{e^{ - n}  + e^{ - 3n} }}<br />
{{1 - e^{ - 2n} }}
    Is the answer now clear?
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