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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$\begingroup$$\ln z$ is the derivative of $t\mapsto z^t$ at $t=0$, so $$\ln z = \lim_{h\to 0}\frac{ z^h-1}h=\lim_{\omega\to \infty} \omega(z^{1/\omega}-1).$$
$\endgroup$ $\begingroup$You have $z^{1/\omega}= \exp ( \ln(z)/\omega)= 1+ \ln(z)/\omega + o(1/\omega)$, so $\ln(z)=\lim\limits_{\omega \to + \infty} \omega (z^{1/\omega}-1)$.
$\endgroup$ 1 $\begingroup$Another way is to use L'Hôpital's rule. Write $w(z^{1/w}-1)$ as $$\frac{z^{1/w}-1}{1/w}\,.$$Note that both the numerator and the denominator go to 0 as $w\to\infty$; thus, we can apply L'Hôpital's rule:$$\lim_{w\to\infty}\frac{z^{1/w}-1}{1/w} =\lim_{w\to\infty}\frac{\frac{d}{dw}(z^{1/w}-1)}{\frac{d}{dw}1/w}=\lim_{w\to\infty}\frac{(-1/w^{2})(\ln z)z^{1/w}}{-1/w^2}=\ln z \,. $$In the numerator, we used $z^{1/w}=e^{(1/w)\ln z}$ and then proceeded as follows:$$ \frac{d}{dw}e^{(1/w)\ln z}=\left[\frac{d}{dw}\Big((1/w)\ln z\Big)\right]e^{(1/w)\ln z}=\left[(-1/w^{2})\ln z\right]z^{1/w}\,. $$
$\endgroup$ $\begingroup$Not a formal proof, only a bit of intuition. Substituting $x=\ln(z)$ (therefore $z=\exp(x)$), and $n$ instead of $\omega$, this limit is aligned with the infamous Euler's limit for the exponential.
$$ x = \lim_{n \to \infty} n (\exp(x) ^{1/n} - 1) = \lim_{n \to \infty} n (\exp(x/n) - 1 $$
We put the limit aside, and use some trivial algebraic manipulations. Then we put the limit back, to get Euler's limit:
$$\exp(x) = \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n$$
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