What's the lim of (e^3t-1)/t, as t aproaches 0?

Question

Problem says to use L'Hopital's rule if it applies, can someone explain this? Thanks!

Answer 1

L'Hopital's rule does apply in this case because the original function is undefined, but it's derivative would not be undefined.

f(t) = e^3t - 1
f'(t) = 3e^3t

g(t) = t
g'(t) = 1

lim t -> 0 (3e^3t) = 3e^0 = 3

Answer 2

lim t→0 (e^3t - 1)/t

e^(3(0)) - 1 = 1-1 = 0, and t = 0, so we have
0/0, so use l'Hopital's rule and find
lim t→0 d[(e^3t - 1)/t]/dt / dt/dt =
lim t→0 3e^3t = 3

Is there a problem with parentheses?

Answer 3

sorry philo was right, read the brackets wrongs.