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

1 Answer
Bdub
Apr 28, 2016

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#

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