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The test is very simple for the case $f(x) = x = g(x)$ since $$ \lim_{x\to 0}{x^x} = 1 $$ But in other cases?

Note that they do not specify that the functions are differentiable and neither that they are continuous throughout their domains. We can assume that the functions are defined in an open interval containing zero.

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  • $\begingroup$ I´m not sure that the proposition is a theorem. Maybe it is false. $\endgroup$ Nov 17, 2014 at 23:52
  • $\begingroup$ It is false. This is the reason why we don't just define 0^0 to be 1. I will try and find the counterexample $\endgroup$ Nov 17, 2014 at 23:53
  • $\begingroup$ Presumably $f(x)$ is non-negative $\endgroup$
    – Henry
    Nov 17, 2014 at 23:53
  • $\begingroup$ Works with $f=g$ positive. $\endgroup$ Nov 18, 2014 at 8:00

3 Answers 3

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Note that $$ \log (f(x)^{g(x)}) = g(x)\log f(x) $$ Now the result holds as soon as $ g(x)\log f(x) \to 0$, but is is not always the case (if $g(x)$ goes to $0$ slow enough).

For instance: $$ f(x) = e^{-1/|x|}\implies g(x)\log f(x) = -\frac {g(x)}{|x|} $$ Now take $g(x) = \sqrt{|x|}$ and this is a counterexample.

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  • $\begingroup$ But f(x) can't be e^(-1/|x|) because in this case f(0) is not 0. $\endgroup$ Nov 17, 2014 at 23:59
  • $\begingroup$ there is a limit in 0. If you define $f(0)=0$ you have a continuous function. $\endgroup$
    – mookid
    Nov 18, 2014 at 0:01
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As a counter-example, consider $f(x)=0$ and $g(x)=x^2$ which are each continuous and differentiable

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  • $\begingroup$ f(x)=0 is not a very interesting counterexample. $\endgroup$
    – mookid
    Nov 17, 2014 at 23:56
  • $\begingroup$ @mookid: but finding its limit and positive powers are easy. $\endgroup$
    – Henry
    Nov 17, 2014 at 23:57
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To prove that $$ \lim_{x\to 0}f(x)^{g(x)}=1 $$ We can instead prove $$ \lim_{x\to 0} {g(x)}\ln(f(x))=0\\\\ $$ We can use L'Hopital's rule to prove this if we rearrange the limit first. $$ \begin{align} &\lim_{x\to 0} g(x)\ln(f(x))\\\\ =&\lim_{x\to 0}\frac{\ln(f(x))}{1/g(x)}\\\\ =&\lim_{x\to 0}\frac{\frac{1}{f(x)}f^\prime(x)}{-\frac{1}{g^2(x)}g^\prime(x)}\\\\ =&\lim_{x\to 0}\frac{g^2(x)}{f(x)}\frac{f^\prime(x)}{g^\prime(x)}\\\\ =&\lim_{x\to 0}g(x)=0 \\\\ \end{align} $$

We can use L'Hopital's rule above since the limit $x\to 0$ of $\frac{f(x)}{g(x)}$ is of the indeterminate form $\frac00$. Since we've proven that the limit of $g(x)\ln(f(x))$ as $x\to 0$ is $0$, we can see that: $$ \lim_{x\to 0}\ln f(x)^{g(x)}=0\implies\lim_{x\to 0}e^{\ln f(x)^{g(x)}}=1\implies\lim_{x\to 0}f(x)^{g(x)}=1 $$

Hopefully this helps answer your question.

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  • $\begingroup$ $f$ positive is required. Even worse, $\lim\frac{g^2f'}{fg'}=\lim g$ is false. $\endgroup$ Nov 18, 2014 at 7:59
  • $\begingroup$ According to L'Hopital's theorem, it holds that $\lim_{x\to 0}\frac{f}{g}=\lim_{x\to 0}\frac{f^\prime}{g^\prime}$. Therefore, shouldn't this work: $\lim\frac{g^2f'}{fg'}=\lim\frac{g^2f}{fg}=\lim g$ The last two examples on this website also relate to the question, along with the "key concepts" section : math.hmc.edu/calculus/tutorials/lhopital $\endgroup$
    – Dr C
    Nov 18, 2014 at 15:04

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