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Natural Logarithm Laws

  1. Nov 16, 2009 #1
    How does

    A friend told me this was a true statement but could'nt prove it. If that isn't true, then how would you find the lim x->0 of (ln(x))^(1/x) using L'Hospital's Rule?
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  3. Nov 16, 2009 #2


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    It doesn't, in general. It does if x = 1.
  4. Nov 16, 2009 #3


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    For example, if x= 2, ln(2)= 0.69315, approximately so [itex](ln(2))^{1/2}= 0.83255[/itex]. But [itex]2^{1/2}= 1.41421[/itex] so [itex]ln(2^{1/2})= 0.34657. Not at all the same.
  5. Nov 16, 2009 #4


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    For example, if x= 2, then ln(2)= 0.69315, approximately, and [itex](ln(2))^{1/2}= 0.83255.

    But [itex]2^{1/2}= 1.41421[/itex] and so [itex]ln(2^{1/2})= 0.34657. Not at all the same.

    As for the entire problem of finding the limit, as x goes to 0, of [itex](ln(x))^{1/x}[/itex], I see a serious difficulty: as soon as x< 1, ln(x)< 0 and fractional powers of negative numbers are not defined.
  6. Nov 16, 2009 #5
    That's what I thought, but my friend insisted that it was true. I've been rattling my brain for about 2 days on that one, so I decided to ask here. So can you help me with limit I mentioned in my first post? I typed it wrong in the first post its the limit as x approaches infinity not zero. I can see from graphing it that it's going to come out to one, but I don't know how to use L'Hopistal's Rule to solve for it. If I try to evaluate it without changing anything I get something like [tex]\infty0[/tex] which would be one if it isn't indeterminant, I don't remember if it is or isn't. Anyways, I'm supposed to use L'Hosp. Rule and I don't know how to write the limit as a fraction to use L'Hopistal's Rule though.
  7. Nov 16, 2009 #6


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    General formula: ln(ab)=(b)ln(a)
    For your formula: ln(x1/x)=(1/x)ln(x)

    As for the L'Hopital rule question, you don't need it, since the expression goes to (-∞), which is ∞, with an ambiguous sign.
    Last edited: Nov 16, 2009
  8. Nov 16, 2009 #7
    The original equation is [ln(x)]^(1/x) not ln(x^(1/x)).
  9. Nov 17, 2009 #8
    ...and the original equation was incorrect, so mathman gave something correct.
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