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Wolfram's website lists this as a limit representation of the natural log:

$$\ln{z} = \lim_{\omega \to \infty} \omega(z^{1/\omega} - 1)$$

Is there a quick proof of this? Thanks

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It is quite interesting that this should come up just now. I used this very limit in this answer last night. –  robjohn Jan 20 '13 at 1:24

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up vote 7 down vote accepted

$\ln z$ is the derivative of $t\mapsto z^t$ at $t=0$, so $$\ln z = \lim_{h\to 0}\frac{ z^h-1}h=\lim_{\omega\to \infty} \omega(z^{1/\omega}-1).$$

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You have $z^{1/\omega}= \exp ( \ln(z)/\omega)= 1+ \ln(z)/\omega + o(1/\omega)$, so $\ln(z)=\lim\limits_{\omega \to + \infty} \omega (z^{1/\omega}-1)$.

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Yes, of course...as long as a definite branch of the complex logarithm is implicitly chosen and used. In this case it seems to be the "usual" branch which is used ("erase" the non-positive real axis, taking zero as the argument of the whole positive real axis). Otherwise the first equality can easily be false. –  DonAntonio Jan 20 '13 at 2:46

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