Question

How do you evaluate `(e^x - 1 - x )/ x^2` as x approaches 0?

Answer 1

1/2

Explanation:

`lim_(x->0) (e^x -1 -x)/x^2 = (e^0 -1 -0)/0 = (1-1)/0 = 0/0`

This is an indeterminate type so use l'Hopital's Rule. That is, take the derivative of the top and the bottom and then find the limit of its quotient.

`lim_(x->0) (e^x-1)/(2x) =( e^0 -1)/0 = (1-1)/0 = 0/0` This is still an indeterminate form so let's use l'Hopital's Rule again.

`lim_(x->0) e^x /2 = e^0 /2 = 1/2`