Posted by Anonymous on Saturday, November 8, 2008 at 4:53pm.
ln(x+h)-lnx = ln[1 + (h/x)=
-> h/x for x ->0
Divide that by h and you get 1/x. The limit as x->0 is infinity
I don't see how to use
limit of (e^h - 1)/h = 1 as h approaches 0 to solve this
e^h -1 -> h + h^2/2! + ... as h-> 0
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