Find the limit?

Question

lim t->0 (e^t - 1) / t^3

This produces the indeterminate form of 0/0, so l'hospital's rule is performed:

lim t->0 e^t / 3t^2

This produces the form 1/0. What can I do to determine the limit from here? Or is there a way to evaluate the limit without using l'hopital's rule at all?

Answer 1

If you reach the form 1/0, the know the limit is either undefined (+∞ or -∞) or does not exist at all - any of these cases is possible.
In this particular case the limit is +∞. One way of seeing this is to use the Taylor expansion for e^t:
lim t->0 ((e^t - 1) / t^3)
= lim t->0 ((1 + t + t^2/2 + ... + t^n/n! + ...) - 1) / t^3
= lim t->0 (t + t^2/2 + ... + t^n/n! + ...) / t^3
= lim t->0 (1 + t/2 + ...) / t^2
Since t^2 -> 0 from above, on both sides of the limit, the limit is +∞.

Answer 2

1/0 is positive infinity (+∞). That IS the limit of the expression!

In fact if you plot the function you will see that it tends towards infinity (from the left and from the right) around t=0.

Answer 3

is it e to the power of t ?
not e to the power of t-1 ?

lim t->0 t*e^(t-1) / 3t
lim t->0 e^(t-1)/3 which is > 0

Answer 4

You keep applying l'Hopital's rule until you get something useful. If you bottom out at a division by zero, then you have to find a different method.

Answer 5

You have performed L'Hospital's rule correctly. The limit is infinity.