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Math Help - Another Limit Problem

  1. #1
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    Another Limit Problem

    Just wondering if somebody can help out with the workings for the following:- lim n-> inf (ln n)^1/n
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    I would say \lim_{n \to \infty}(\ln(n))^{\frac{1}{n}} =(\ln(\infty))^{\frac{1}{\infty}} =(\ln(\infty))^{0}=1
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    Thanks for that. So bloody obvious.
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    Quote Originally Posted by p75213 View Post
    Just wondering if somebody can help out with the workings for the following:- lim n-> inf (ln n)^1/n
    (\ln{n})^{\frac{1}{n}} = e^{\ln{[(\ln{n})^{\frac{1}{n}}]}}

     = e^{\frac{1}{n}\ln{(\ln{n})}}

     = e^{\frac{\ln{(\ln{n})}}{n}}.


    Therefore:

    \lim_{n \to \infty}(\ln{n})^{\frac{1}{n}} = \lim_{n \to \infty} e^{\frac{\ln{(\ln{n})}}{n}}

     = e^{\lim_{n \to \infty}\frac{\ln{(\ln{n})}}{n}}

     = e^{\lim_{n \to \infty}\frac{\frac{1}{n\ln{n}}}{1}} by L'Hospital's Rule

     = e^{\lim_{n \to \infty}\frac{1}{n\ln{n}}}

     = e^0

     = 1.
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  5. #5
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    Quote Originally Posted by pickslides View Post
    I would say \lim_{n \to \infty}(\ln(n))^{\frac{1}{n}} =(\ln(\infty))^{\frac{1}{\infty}} =(\ln(\infty))^{0}=1
    Actually \infty^0 is an indeterminate form.

    You need to undergo a transformation to get something of the form \frac{0}{0} or \frac{\infty}{\infty} so you can use L'Hospital's Rule.
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