Question

How do you evaluate `(e^(7t)-1)/(t)` as t approaches 0?

Answer 1

Limit is 7.

Explanation:

We have indeterminate limit at first, ie `0/0`. So apply L'Hopital's rule:

`lim_(t->0) (e^(7t)-1)/t = lim_(t->0) (d/(dt)(e^(7t)-1))/(d/(dt)(t))`

`=lim_(t->0) (7e^(7t))/1 = 7`