calculus
posted by Anonymous .
use the rule that says
limit of (e^h  1)/h = 1 as h approaches 0
to show that the limit of [ln(x+h) lnx]/h as h approaches 0 = 1/x, where x>0

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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