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Math Help - find this limit

  1. #1
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    find this limit

    limit ( n* ( a^(1/n)-1), n tends to infinity)

    Thx, mate!
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  2. #2
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    Quote Originally Posted by szpengchao View Post
    limit ( n* ( a^(1/n)-1), n tends to infinity)
    Substitute u=\frac1n,

    \mathop {\lim }\limits_{n \to \infty } n\left( {a^{1/n}  - 1} \right) = \mathop {\lim }\limits_{u \to 0} \frac{{a^u  - 1}}<br />
{u},

    and this is fairly well known, the answer is \ln a for a>0. Substitute v=\frac1{a^u-1}.
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  3. #3
    GAMMA Mathematics
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    Quote Originally Posted by Krizalid View Post
    and this is fairly well known, the answer is \ln a for a>0. Substitute v=\frac1{a^u-1}.
    Just in case you were wondering why it is natural log, let me show you using L'Hopital's Rule.

    When the limit ends up being in the form of \frac{\infty}{\infty} OR \frac{0}{0}, then take the derivative of the top over the derivative of the bottom: \frac{f'(x)}{g'(x)} and take the limit...

    lim \frac{a^u-1}{u} = lim \frac{a^u ln|a|}{1} \Rightarrow ln|a|
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  4. #4
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    Quote Originally Posted by colby2152 View Post
    Just in case you were wondering why it is natural log, let me show you using L'Hopital's Rule.
    The final substitution I left there provides a solution without L'H˘pital.
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