Date: Apr 27, 2001 5:55 PM
Author: S.K.Mody
Subject: Re: A limit for ln(x)?




"Marko" <kroatman@yahoo.se> wrote in message
news://3AE9D37D.D138C124@yahoo.se...
> This limit seems to give ln(x):
>
> lim n*(x^1/n - 1) = ln(x), x > 0.
> n->oo
>


Are you familiar with the limit:-
exp(x) = lim[n -> oo](1 + x/n)^n ----- (1)

To show this, fix an x in R and expand (1) in a binomial
series and you get a power series which approximates the
series for exp(x).

Inverting (1) gives the limit you mention.
Let y(n) = (1 + x/n)^n
and y = exp(x)

Then:
ln(y) = x = n*( y(n)^(1/n) - 1 )
= n*( (y + y(n) - y)^(1/n) - 1 )
= n*( y^(1/n) - 1 ) + O(y - y(n))
= x(n) + O(y - y(n))

Regards,
S.K.Mody.