Trying to show that $\ln(x) = \lim_{n\to\infty} n(x^{1/n} -1)$ How do I show that $\ln(x) = \lim_{n\to\infty} n (x^{1/n} - 1)$?
I ran into this identity on this stackoverflow question. I haven't been able to find any proof online and my efforts to get from $\ln(x) := \int_1^x \frac{\mathrm dt}t$ to that limit have been a failure.

You can even do a bit more using Taylor series $$x^{\frac 1n}=e^{\frac 1 n \log(x)}=1+\frac{\log (x)}{n}+\frac{\log ^2(x)}{2
   n^2}+O\left(\frac{1}{n^3}\right)$$ which makes $$n(x^{\frac 1n} -1)=\log (x)+\frac{\log ^2(x)}{2 n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and also how it is approached.
