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Ln limit

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data
    find the following limit: ln((2x)^(2x))/(3x²+1) with x>infinity.

    3. The attempt at a solution
    I tried the following: lim ( ln((2x)^(2x))/(3x²+1) ) with x>infity =
    lim ( (3x²+1)/ln((2x)^(2x)) ) with x>0
    lim 2xln(2x) with x>0 = 0, therefore lim ( (3x²+1)/(2xln(2x)) with x>0 =
    1/0 which is infinity, a simple plot however shows this is wrong, could someone help me?
    Last edited: Oct 11, 2009
  2. jcsd
  3. Oct 11, 2009 #2


    Staff: Mentor

    The step above isn't valid. There is no theorem that says that
    [tex]\lim_{x \rightarrow \infty} \frac{a}{b}~=~\lim_{x \rightarrow 0}\frac{b}{a}[/tex]
    which is what it seems you have done. IOW, you can't just flip the numerator and denominator and change the limit variable.

    As x gets large without bound, both numerator and denominator also get large without bound. The simplest way to do this, I believe, would be to use L'Hopital's Rule.
  4. Oct 11, 2009 #3
    okay, thank you.
    I am, however, not allowed to use L´Hopital´s rule..
    does anyone know a method besides L´Hopital´s rule?
  5. Oct 11, 2009 #4


    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi eric123! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    Hint: How fast does ln(x) go up, compared with x? :smile:
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