Roger Smith

2022-01-03

Trying to show that
$\mathrm{ln}\left(x\right)=\underset{n\to \mathrm{\infty }}{lim}n\left({x}^{\frac{1}{n}}-1\right)$

alexandrebaud43

$\underset{n\to \mathrm{\infty }}{lim}n\left({x}^{\frac{1}{n}}-1\right)=\underset{n\to \mathrm{\infty }}{lim}\frac{{x}^{\frac{1}{n}}-1}{\frac{1}{n}}={f}^{\prime }\left(0\right)$ , where $f\left(t\right)={x}^{t}$. Since
${f}^{\prime }\left(t\right)=\mathrm{ln}\left(x\right){x}^{t}$
it follows that ${f}^{\prime }\left(0\right)=\mathrm{ln}\left(x\right)$

Bubich13

Set $x={e}^{t},$ then
$\underset{n\to \mathrm{\infty }}{lim}n\left({x}^{\frac{1}{n}}-1\right)=t\underset{n\to \mathrm{\infty }}{lim}\frac{{e}^{\frac{t}{n}}-1}{\frac{t}{n}}$
$=t\underset{u\to 0}{lim}\frac{{e}^{u}-1}{u}$
$=t$
$=\mathrm{log}\left(x\right)$

Vasquez

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

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