# calculus

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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

• calculus -

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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