Question

given e^x=1+x+x^2/2!+x^3/3!+... how do you prove that lim((e^(h)-1)/h)=1 as x->0

Answer 1

This looks like a question posted yesterday. Was it by you?

My answer is still the same.

Write out e^h using the series you were given. (The x becomes h, but it's the same series.) Then subtract 1. Divide what is left by h. You should be left with a series that starts with 1 and then has a sum of difference powers of h. Those h^n terms become zero then h->0, leaving you with 1.

Answer 2

i don't know why but i keep getting 0.
i first take the 1 out of the inputed series and subtract it from the other 1 to get (x+x^2/2!+...)/x. then i take an x out and divide it by the x in the denominator to get x/2!+x^2/3!+... but when you plug 0 into that all you get is 0.

Answer 3

Here is where you went wrong:

<< then i take an x out and divide it by the x in the denominator to get x/2!+x^2/3!+... but when you plug 0 into that all you get is 0. >>

You don't take an x out. When you divide
x + x^2/2!+x^3/3!+... by x you get
1 + x/2! + x^2/3! + ...