Power series of e

1. Mar 28, 2006

beanryu

Use a four-term Taylor approximation for e^h , for h near 0 , to evaluate the following limit.

lim (e^h-1-h-h^2/2)/h^3
h->0

i know that e^h = 1+h+h^2/2+h^3/3+h^4/4......

therefore, I say that e^h-1-h-h^2/2 = h^3/3+h^4/4........

(h^3/3+h^4/4.....)/h^3 is approximately = 1/3

but its wrong

please give me some hints THANX!!!

2. Mar 28, 2006

Tom Mattson

Staff Emeritus
It's not wrong.

3. Mar 29, 2006

beanryu

But It Is....

4. Mar 29, 2006

Tom Mattson

Staff Emeritus
Ack, you're right it is wrong. Your mistake is in the terms of the series.

Nope, those denominators should have factorials in them, like so:

e^h = 1+h+h^2/2!+h^3/3!+h^4/4!
e^h = 1+h+h^2/2+h^3/6+h^4/24

That should fix it.