L'hopitals rule?

Question

Use L'Hopital's rule to solve the following

lim ((e^(x) - 1) / x)
x->0

Answer 1

Hi! Your problem there is a classic l'hopital problem.

You can use L'hoptial's rule when your expression tends to 0/0 or infinity over infinity. In your case you have zero over zero.

L'hopital's rule says that if you have any of the above, you can take the derivative of top and bottom, and then evaluate that new expression for your limit.

(e^x - 1) over x --- take the derivative of the top and bottom

(e^x)/1. Now evaluate that at your limit, zero.

e^0 = 1.

Hope that helped!!
UMRmathmajor.

Answer 2

[x→0]lim (e^x - 1)/x = [x→0]lim e^x = 1