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September 1, 2014

September 1, 2014

Posted by **Anonymous** on Saturday, November 8, 2008 at 4:53pm.

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 -
**drwls**, Saturday, November 8, 2008 at 8:59pmln(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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