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Confusing log limit

  • #1
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Homework Statement



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

Homework Equations


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

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?
 

Answers and Replies

  • #2
Math_QED
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Homework Statement



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

Homework Equations


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

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?
0^0 is undefined
 
  • #3
RUber
Homework Helper
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344
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.
 
  • #4
Ray Vickson
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Homework Statement



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

Homework Equations


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

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?
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.
 
  • #5
Math_QED
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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.
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).
 
  • #6
Ray Vickson
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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).
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).
 
  • #7
Math_QED
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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).
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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