Roger Smith

2022-01-03

Trying to show that

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

alexandrebaud43

Beginner2022-01-04Added 36 answers

it follows that

Bubich13

Beginner2022-01-05Added 36 answers

Set $x={e}^{t},$ then

$\underset{n\to \mathrm{\infty}}{lim}n({x}^{\frac{1}{n}}-1)=t\underset{n\to \mathrm{\infty}}{lim}\frac{{e}^{\frac{t}{n}}-1}{\frac{t}{n}}$

$=t\underset{u\to 0}{lim}\frac{{e}^{u}-1}{u}$

$=t$

$=\mathrm{log}\left(x\right)$

Vasquez

Expert2022-01-09Added 669 answers

You can even do a bit more using Taylor series

which makes

which shows the limit and also how it is approached.

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