# Trying to show that \ln(x)=\lim_{n\rightarrow\infty}n(x^{\frac{1}{n}}-1)

Trying to show that
$$\displaystyle{\ln{{\left({x}\right)}}}=\lim_{{{n}\rightarrow\infty}}{n}{\left({x}^{{{\frac{{{1}}}{{{n}}}}}}-{1}\right)}$$

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alexandrebaud43
$$\displaystyle\lim_{{{n}\rightarrow\infty}}{n}{\left({x}^{{{\frac{{{1}}}{{{n}}}}}}-{1}\right)}=\lim_{{{n}\rightarrow\infty}}{\frac{{{x}^{{{\frac{{{1}}}{{{n}}}}}}-{1}}}{{{\frac{{{1}}}{{{n}}}}}}}={f}'{\left({0}\right)}$$ , where $$\displaystyle{f{{\left({t}\right)}}}={x}^{{t}}$$. Since
$$\displaystyle{f}'{\left({t}\right)}={\ln{{\left({x}\right)}}}{x}^{{t}}$$
it follows that $$\displaystyle{f}'{\left({0}\right)}={\ln{{\left({x}\right)}}}$$
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Bubich13
Set $$\displaystyle{x}={e}^{{t}},$$ then
$$\displaystyle\lim_{{{n}\rightarrow\infty}}{n}{\left({x}^{{{\frac{{{1}}}{{{n}}}}}}-{1}\right)}={t}\lim_{{{n}\rightarrow\infty}}{\frac{{{e}^{{{\frac{{{t}}}{{{n}}}}}}-{1}}}{{{\frac{{{t}}}{{{n}}}}}}}$$
$$\displaystyle={t}\lim_{{{u}\rightarrow{0}}}{\frac{{{e}^{{u}}-{1}}}{{{u}}}}$$
$$\displaystyle={t}$$
$$\displaystyle={\log{{\left({x}\right)}}}$$
Vasquez

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