Question

How do you find the limit of `( e^(3t) - 1 ) / t` as x approaches 0?

Answer 1

`3`

Explanation:

We will be using `e^x= sum_{n=0}^{infty}x^n/(n!)`
Taking `e^{3t} = 1 + 3t+(3t)^2/2+(3t)^3/6+...` and substituting
`lim_{t->0}(e^{3t}-1)/t = lim_{t->0}((1+ 3t+(3t)^2/2+(3t)^3/6+...-1)/t)`
`lim_{t->0}(e^{3t}-1)/t =lim_{t->0}(t((3+(3t)/2+(3t)^2/6+...))/t)=3`