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Math Help - limit e problem

  1. #1
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    limit e problem

    This is similar to something I posted yesterday, but I can't figure it out.

    What's the limit of the following as n goes to infinity,

    [(a^(1/n) + b^(1/n))/2]^n for a, b > 0

    ?

    Could someone show me how to do this with algebra?
    Thanks in advance.
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  2. #2
    MHF Contributor red_dog's Avatar
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    This is the case 1^{\infty}.

    \displaystyle\lim_{n\to\infty}\left(\frac{a^{\frac  {1}{n}}+b^{\frac{1}{n}}}{2}\right)^n=\lim_{n\to\in  fty}\left(1+\frac{a^{\frac{1}{n}}+b^{\frac{1}{n}}-2}{2}\right)^n=e^{\displaystyle\lim_{n\to\infty}n\  left(\frac{a^{\frac{1}{n}}+b^{\frac{1}{n}}-2}{2}\right)}=

    \displaystyle =e^{\displaystyle\frac{1}{2}\lim_{n\to\infty}\left  (\frac{a^{\frac{1}{n}}-1}{\frac{1}{n}}+\frac{b^{\frac{1}{n}}-1}{\frac{1}{n}}\right)}=e^{\frac{1}{2}(\ln a+\ln b)}=e^{\ln\sqrt{ab}}=\sqrt{ab}
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  3. #3
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    Quote Originally Posted by BrainMan View Post
    This is similar to something I posted yesterday, but I can't figure it out.

    What's the limit of the following as n goes to infinity,

    [(a^(1/n) + b^(1/n))/2]^n for a, b > 0

    ?

    Could someone show me how to do this with algebra?
    Thanks in advance.
    Use the fact that,
    \left( 1 + \frac{s_n}{n} \right)^n \to e^s where s_n \to s.

    So here, 2s_n = na^{1/n} + nb^{1/n} - 2n

    This means, 2s_n = n(a^{1/n} - 1) + n(b^{1/n} - 1).
    Thus,
    2s = \log a + \log b = \log ab.

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