Question

How do you evaluate the limit `lim (e^t-1)/sint` as `t->0`?

Answer 1

Currently we're in indeterminate form, `0/0`, so we can use l'Hopitals rule to solve the problem.

`lim_(x-> c) (f(x))/(g(x)) = lim_(x-> c) (f'(x))/(g'(x))`

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

`=e^0/cos(0)`

`= 1/1`

`= 1`

Hopefully this helps!