L'hopital's Rule, help please?

lim t->0 (e^t - 1) / t^9

L'Hopitals rule apply since both the numerator and denominator are 0. Take the derivatives of both the numerator and denominator separately:

= lim t->0 (e^t) / (9t^8)

Now l'Hopital no longer applies since e^0 = 1, and 9*0^8 = 0.

The limit does not exist.

L'hopital's rule indicates that you can take derivatives of the numerator and denominator (possibly multiple times) in order to evaluate a limit. The derivative of e^x is e^x. Using this information, we can evaluate the limit as follows:

f(x) = (e^t - 1) / t^9

Applying L'hopital's rule successively gives

e^t / 9t8

e^t / 72t^7

e^t / 504t^6

e^t / 3024t^5

e^t / 15120t^4

e^t / 60480t^3

e^t / 181440t^2

e^t / 362880t

e^t / 362880

Now, as we take the limit as t->0, the limit is 1 / 362880.