Question

how do you show that lim((e^(h)-1)/h)=1 as h approches 0 given the series e^x= 1+x+(x^2/2!)+(x^3/3!)...
And given F(x)=e^x how do you find the lim((f(x+h)-f(x))/h)as h approches 0

Answer 1

One problem at a time, please.

Using the series form of e^h:
(e^h -1)/h = 1 + h/2! + h^2/3! + ... h^n/(n+1)!
As h--> 0, all of the terms that involve h become zero and only the 1 remains. Therefore that is the limit.

Answer 2

2. That would be the definition of f'(x), which would be e^x = 1.
One can arrive at that result using the series expansion.

lim((f(x+h)-f(x))/h)
x-->0
= (1/h) Lim [1 + (x+h) + (x+h)^2/2! ... - 1 -x -x^2/2! - ...]/h
= Lim 1 + x + (1/3!)[(x+h)^3-x^3] + ...
= 1 + x + x^2/2 + ...
= e^x