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Confusing log limit

  1. Jun 9, 2016 #1
    1. The problem statement, all variables and given/known data

    [tex]\lim\limits_{x \to 0} \left(\ln(1+x)\right)^x[/tex]

    2. Relevant equations
    Maclaurin series:
    [tex]\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + ... + (-1)^{r+1} \frac{x^r}{r} + ...[/tex]

    3. The attempt at a solution

    We're considering vanishingly small [itex]x[/itex], so just taking the first term in the Maclaurin series the limit becomes:

    [tex]\lim\limits_{x \to 0} \left(\ln(1+x)\right)^x = \lim\limits_{x \to 0} x^x = \mathrm{undefined}[/tex]

    or so I thought until google tells me that [itex]0^0 = 1[/itex].

    What's going on here? How can I evaluate the limit properly?
     
  2. jcsd
  3. Jun 9, 2016 #2

    Math_QED

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    0^0 is undefined
     
  4. Jun 9, 2016 #3

    RUber

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    Because ##0^0## is undefined, you have to take the limit, as was given in the problem.
    ##\lim_{x\to 0} (\ln(1+x))^x ## can be looked at as a related rates sort of problem...is the log part going to zero faster then the exponent is taking it back out to 1?
    In either case, the expected options for the solution should be 0 or 1.
    Let's look at this as a sequence:
    ##\{x_n\} = [ \ln ( 1 + \frac{1}{2^n}) ]^ {\frac{1}{2^n} }##
    ##\lim_{x\to 0} (\ln(1+x))^x = \lim_{n\to \infty} x_n##
    For n = 1, ##x_n= \ln(2)##
    Now check the ratio ##\frac{x_n}{x_{n+1}}##. If this is >1, then the sequence is shrinking...if it is <1, then the sequence is increasing.
     
  5. Jun 9, 2016 #4

    Ray Vickson

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    Let ##F(x) = \left( \ln (1+x) \right)^x##. Then ##L(x) \equiv \ln F(x)## is given by
    [tex] \begin{array}{rcl}
    L(x) &=& x \ln \left( \ln (1+x) \right) = x \ln( x - x^2/2 + x^3/3 - \cdots) \\
    &=& x \ln x + x \ln ( 1 - x/2 + x^2/3 - \cdots) .
    \end{array} [/tex]
    Thus, ##\lim_{x \to 0} L(x) = \lim_{x \to 0} x \ln x ##. You can write
    [tex] \lim_{x \to 0} x \ln x = \lim_{x \to 0} \frac{\ln x}{1/x}, [/tex]
    and evaluate that last form using l'Hospital's rule.
     
  6. Jun 11, 2016 #5

    Math_QED

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    Am I missing something because I think you can't apply l'Hospital's rule because lim x>0 lnx is not defined? (We can't approach lnx from the left as it is not defined in the context of real numbers).
     
  7. Jun 11, 2016 #6

    Ray Vickson

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    The original limit cannot be taken from the left, either. When ##x < 0##, ##\ln(1+x) < 0## and ##(\ln(1+x))^x## is undefined for fractional negative ##x## (being a negative raised to a negative fractional power).
     
  8. Jun 11, 2016 #7

    Math_QED

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    Well you could say that the limit does not exist then, but the OP might not have mentioned that they meant the right side limit.
     
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