Evaluate Limit Using Taylor Approximation of Power Series of e^h

[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!

sorry for posting it at precalculus... please help urgent!

 

[quantumdude]

It's not wrong.

 

[beanryu]

But It Is...

 

[quantumdude]

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

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

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.