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

Roger Smith 2022-01-03 Answered
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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Expert Answer

alexandrebaud43
Answered 2022-01-04 Author has 102 answers
\(\displaystyle\lim_{{{n}\rightarrow\infty}}{n}{\left({x}^{{{\frac{{{1}}}{{{n}}}}}}-{1}\right)}=\lim_{{{n}\rightarrow\infty}}{\frac{{{x}^{{{\frac{{{1}}}{{{n}}}}}}-{1}}}{{{\frac{{{1}}}{{{n}}}}}}}={f}'{\left({0}\right)}\) , 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
Answered 2022-01-05 Author has 4448 answers
Set \(\displaystyle{x}={e}^{{t}},\) then
\(\displaystyle\lim_{{{n}\rightarrow\infty}}{n}{\left({x}^{{{\frac{{{1}}}{{{n}}}}}}-{1}\right)}={t}\lim_{{{n}\rightarrow\infty}}{\frac{{{e}^{{{\frac{{{t}}}{{{n}}}}}}-{1}}}{{{\frac{{{t}}}{{{n}}}}}}}\)
\(\displaystyle={t}\lim_{{{u}\rightarrow{0}}}{\frac{{{e}^{{u}}-{1}}}{{{u}}}}\)
\(\displaystyle={t}\)
\(\displaystyle={\log{{\left({x}\right)}}}\)
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Vasquez
Answered 2022-01-09 Author has 8850 answers

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.

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