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Math Help - Limit

  1. #1
    odx
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    [SOLVED] Limit

    Hello
    I'm having difficulties with this limit:

    \lim_{x\rightarrow e^{-}}(ln x)^{\frac{1}{ln^{2}ln x}}


    Thanks for help
    Last edited by odx; December 22nd 2011 at 07:13 PM.
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  2. #2
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    Re: Limit

    \displaystyle \begin{align*} \lim_{x \to e^{-}}\left(\ln{x}\right)^{\frac{1}{\left[\ln{\left(\ln{x}\right)}\right]^2}} &= \lim_{x \to e^{-}}e^{\ln{\left\{ \left(\ln{x}\right)^{\frac{1}{\left[\ln{\left(\ln{x}\right)}\right]^2}} \right\}}} \\ &= \lim_{x \to e^{-}} e^{\left\{\frac{1}{\left[\ln{\left(\ln{x}\right)}\right]^2}\right\}\left[\ln{\left(\ln{x}\right)}\right]} \\ &= \lim_{x \to e^{-}} e^{\frac{1}{\ln{\left(\ln{x}\right)}}} \\ &= \infty \end{align*}
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  3. #3
    odx
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    Re: Limit

    That's the same pattern that I followed, but on my book the solution to that limit is 0 ...I'm starting to think that's a typo


    p.s. thanks for the fast reply
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  4. #4
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    Re: Limit

    I know what my mistake was.

    \displaystyle \begin{align*} \frac{1}{X} \to -\infty \end{align*} as \displaystyle \begin{align*} X \to 0 \end{align*} from the left.

    So \displaystyle \begin{align*} e^X \to 0 \end{align*} as \displaystyle \begin{align*} X \to -\infty \end{align*}.
    Last edited by Prove It; December 22nd 2011 at 05:29 PM.
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  5. #5
    odx
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    Re: Limit

    That's right, got it Have a good day
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