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Math Help - Limits and Logs

  1. #1
    mei
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    Limits and Logs

    Hey there!

    I'm at a loss on how to solve this particular limit:


    lim (1/(x-1))^(ln x)
    x-> 1+

    I have tried doing ln y = ln (1/(x-1))^(ln x)

    and then I got to

    lim ln x * ln (1/(x-1))
    x->1+

    I'm not quite sure how to get out of there. I tried using L'Hopital's rule but I think I must have done something wrong. I would like to know what's the best approach to solve such limits.

    Thanks in advance for your help!
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  2. #2
    MHF Contributor chisigma's Avatar
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    Setting t=x-1 we obtain...

    \lim_{x \rightarrow 1+} \frac{1}{(x-1)^{\ln x}} = \lim_{t \rightarrow 0+} e^{-\ln t \cdot \ln (1+t)} = \lim _{t \rightarrow 0+} e^{-\ln t \cdot \sum_{n=1}^{\infty} (-1)^{n+1} \frac{t^{n}}{n}} = e^{-0} = 1

    Kind regards

    \chi \sigma
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  3. #3
    mei
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    Thanks but can you explain something else to me? How did the ln(x) manage to get on the bottom part?
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  4. #4
    MHF Contributor
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    Quote Originally Posted by mei View Post
    Thanks but can you explain something else to me? How did the ln(x) manage to get on the bottom part?
    Here's an example - \left(\frac{1}{2}\right)^3 = \frac{1}{2^3} .
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  5. #5
    mei
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    I see... because 1 to the power of anything will always be 1!
    Thank you so much! =)
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