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Math Help - Limit question

  1. #1
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    Limit question

    I need help in calculating this limit:

    lim x->e+ (logx)^(1/x-e)

    This gives me an indeterminate form which is 1^inf. ( log(e)=1; 1/(e-e) = 1/0+ = inf )

    I proceed to follow a process that always helped me in these indeterminate forms, which is applying the exponential:

    lim x->e+ e^[log(logx)^(1/(1-e))] =

    = lim x->e+ e^[(log(logx))/(x-e)]

    Now, it's when I get confused:

    I proceed to substitute the x for the e+, which gives me:

    e^[(log(log(e))/(e-e)]

    log(e) = 1; therefore log(log(e)) = log(1) = 0

    e-e = 0; since it's a lateral limit form the right side, e-e=0+

    Which gives me the result of e^(0/0+)

    I don't know where to go from here. Is 0/0+ and indeterminate form? Did I do something wrong?
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  2. #2
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    If everything is correct, then \frac{0}{0.00000...1}=0

    e^{0}=1
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  3. #3
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    Quote Originally Posted by CMartins View Post
    I need help in calculating this limit:

    lim x->e+ (logx)^(1/x-e)
    let \displaystyle y = \lim_{x \to e^+} (\log{x})^{\frac{1}{x-e}}

    \displaystyle \log{y} = \lim_{x \to e^+} \frac{\log(\log{x})}{x-e}<br />

    L'Hopital ...

    \displaystyle \log{y} = \lim_{x \to e^+} \frac{1}{x\log{x}} = \frac{1}{e}

    \displaystyle y = e^{\frac{1}{e}}
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  4. #4
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    I think, it should be something like this
    lim x->e+ (logx)^(1/(x-e)) = lim x->e+ e^((1/(x-e))*log(log x)).
    By L' Hospital's rule, this is
    im x->e+ e^(1/(x*log(x))) = e^(1/e).
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  5. #5
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by CMartins View Post
    I need help in calculating this limit:

    lim x->e+ (logx)^(1/x-e)

    This gives me an indeterminate form which is 1^inf. ( log(e)=1; 1/(e-e) = 1/0+ = inf )

    I proceed to follow a process that always helped me in these indeterminate forms, which is applying the exponential:

    lim x->e+ e^[log(logx)^(1/(1-e))] =

    = lim x->e+ e^[(log(logx))/(x-e)]

    Now, it's when I get confused:

    I proceed to substitute the x for the e+, which gives me:

    e^[(log(log(e))/(e-e)]

    log(e) = 1; therefore log(log(e)) = log(1) = 0

    e-e = 0; since it's a lateral limit form the right side, e-e=0+

    Which gives me the result of e^(0/0+)

    I don't know where to go from here. Is 0/0+ and indeterminate form? Did I do something wrong?
    Couldn't we do,

     y= (lnx)^{ \frac{1}{x-e}}

     lny = \frac{1}{x-e} ln (lnx)

     \lim_{x \to e^+ } lny = \frac { ln ( lnx ) }{x-e}

    This, of course is indeterminent, so apply Lhopitals to get

     \lim_{x \to e^+ } lny= \frac{ \frac{1}{lnx} \frac{1}{x} }{1}

    Which results in,

      \lim_{x \to e^+ } lny = \frac{1}{e}

    Since this is the limit of lny, the limit of y is simply,

      \lim_{x \to e^+ } y = e^{\frac{1}{e}}

    That seems to agree with wolfram, lim &#40;lnx&#41;&#94;&#40;1&#47;&#40;x-e&#41;&#41;, x&#61;e - Wolfram|Alpha

    I am not very good with limits though! But I think this is good?
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  6. #6
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    Ah! I get it now.
    Thank you.
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