Final answer:
The limit as t approaches 0 of (e^t - 1) / t³ is found to be 1/6 after applying L'Hôpital's Rule three times to resolve the indeterminate form of 0/0.
Step-by-step explanation:
To calculate the limit as t approaches 0 of (e^t - 1) / t³, we must recognize this as an indeterminate form of the type 0/0. Applying L'Hôpital's Rule consecutively, we take derivatives of the numerator and the denominator until we can evaluate the limit directly. Let's go through the steps to calculate this limit. The first derivative of the numerator e^t is just e^t, and the derivative of 1 is 0. The first derivative of t³ is 3t². Applying L'Hôpital's Rule once, we have:
Limit as t approaches 0 of (e^t - 1) / t³ = limit as t approaches 0 of e^t / (3t²).
If we apply L'Hôpital's Rule a second time, the derivative of e^t remains e^t, and the derivative of 3t² is 6t. This gives us:
Limit as t approaches 0 of e^t / (3t²) = limit as t approaches 0 of e^t / (6t).
Applying L'Hôpital's Rule a third time, we find that the numerator is still e^t and the derivative of 6t is 6, leading to the result:
Limit as t approaches 0 of e^t / (6t) = limit as t approaches 0 of e^t / 6.
Finally, substituting t with 0 in the expression e^t / 6, we get:
Limit as t approaches 0 of e^0 / 6 = 1/6.