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Limit of a function

  1. May 13, 2008 #1
    I have no idea of the following question, please help me

    Find limit x→∞〔 (a^x - 1 )/(a - 1) 〕^1/x

    for (1) 0<a<1
    (2) a>1
     
  2. jcsd
  3. May 13, 2008 #2

    malawi_glenn

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    what have you tried? If you are not showing your effort, then you can't get any help
     
  4. May 13, 2008 #3
    let y=〔 (a^x - 1 )/(a - 1) 〕^1/x

    Iny=1/x In(a^x - 1 )/(a - 1)

    then I can't do it
     
  5. May 13, 2008 #4

    malawi_glenn

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    start with 0<a<1 then what happens?
     
    Last edited: May 13, 2008
  6. May 13, 2008 #5
    i really don't know
     
  7. May 13, 2008 #6
    limit x→∞〔 (a^x - 1 )/(a - 1) 〕^1/x

    = [(-1)/(a-1)]^0 = 1

    is it correct?
     
  8. May 13, 2008 #7

    benorin

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    Keep going! Let [tex]x\to\infty[/tex]

    [tex]\lim_{x\to\infty}\frac{\ln\left(\frac{a^x-1}{a-1}\right)}{x}=\frac{\lim_{x\to\infty}\ln\left(\frac{a^x-1}{a-1}\right)}{\lim_{x\to\infty}x}[/tex]

    and remember ln(-1/(a-1)) is some constant.
     
  9. May 13, 2008 #8

    malawi_glenn

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    yes! :-)

    And what if a>1 ?
     
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