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Re: A limit for ln(x)?
Posted:
Apr 27, 2001 5:55 PM


"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.



