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