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Find log_x(4) as x-->1

  1. Oct 25, 2008 #1
    The problem statement, all variables and given/known data
    Evaluate
    lim x-->1 logx(4)


    The attempt at a solution
    I can't understand this because basically if you plug in 1 as x, log14 doesnt have a solution because 1 to the power of anything is just 1.
     
  2. jcsd
  3. Oct 25, 2008 #2

    HallsofIvy

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    Re: Limits

    So that is clearly not a continuous function. But you don't find a limit by "plugging in" a number. If y= logx(4), then 4= xy. I think what I would do is let h= x-1 so that 4= (1+h)y and the limit is as h goes to 0. Apply the generalized binomial formula to (1+ h)y
     
    Last edited: Oct 26, 2008
  4. Oct 25, 2008 #3
    Re: Limits

    Thanks for the response, now I remember that I'm not looking for the value at 1, so that makes sense now. But I'm not sure what to do with that binomial as we haven't covered binomial formula.

    I tried thinking about it with RHS and LHS limits. If you're approaching 1 from the left then y is a large negative and if you are approaching from the right then y is a large positive number, therefore the limit at 1 does not exist. Does that make sense?
     
  5. Oct 25, 2008 #4

    statdad

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    Re: Limits

    HallsofIvy, correct me if I'm off track, but I think your approach hides the dependence of the limit in this variable [tex] y [/tex], and since there is no simple way to evaluate the generalized series, I'm not sure where to turn.

    My first thought was this:

    [tex]
    \log_x {4} = \frac{\log 4}{\log x}
    [/tex]

    (I used logs base 10; obviously [tex] \ln [/tex] would also do)
    Then look at the limit of this expression from the right and left. Explorations with SAGE confirmed my suspicions about this.

    If I've missed something obvious, please let me know.
     
  6. Oct 26, 2008 #5
    Re: Limits

    So even with [tex]

    \log_x {4} = \frac{\log 4}{\log x}

    [/tex] it looks like the limit does not exist because when approaching 1 from the left the denominator is a very small negative number making the limit negative infinity and when approaching from the right the denominator is a very small positive number making the limit postiive infinity, therefore limit does not exist, is that correct?
     
  7. Oct 26, 2008 #6

    Gib Z

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    Re: Limits

    Yes.
     
  8. Oct 26, 2008 #7

    HallsofIvy

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    Re: Limits

    Actually that was my point. Using the generalized Binomial formula for (1+ h)y gives 1+ hy+ higher power terms in h. 4= 1+ hy+ higher power terms in h. Taking the limit as h goes to 0 gives 4= 1 no matter what y is: the limit does not exist.
     
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