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Re: A limit for ln(x)?
Posted:
Apr 27, 2001 5:37 PM


In article <3AE9D37D.D138C124@yahoo.se>, Marko <kroatman@yahoo.se> wrote:
> This limit seems to give ln(x): > > lim n*(x^1/n  1) = ln(x), x > 0. > n>oo
Fixing x > 0, the above is the derivative of x^t wrt t at t = 0, because
n*(x^1/n  1) = (x^1/n  x^0)/(1/n).
Recalling that d(x^t)/dt = ln(x)*x^t, putting in t = 0 gives your result.
Wade



