Calculus of a Single Variable, Chapter 8, 8.7, Section 8.7, Problem 15

Given to solve ,
lim_(x->0^(+)) (e^x - (1+x)) / x^3
as x->0+ then the (e^x- (1+x)) / x^3=0/0 form
so upon applying the L 'Hopital rule we get the solution as follows,
as for the general equation it is as follows
lim_(x->a) f(x)/g(x) is = 0/0 or (+-oo)/(+-oo) then by using the L'Hopital Rule we get the solution with the below form.
lim_(x->a) (f'(x))/(g'(x))

so, now evaluating
lim_(x->0^(+)) (e^x - (1+x)) / x^3
=lim_(x->0^(+)) (e^x - (1+x))' / (x^3)'
= lim_(x->0^(+)) ((e^x - 1)) / ((3x^2))
When x->0+ we get (e^x - 1) /(3x^2) = 0/0 form, so applying the l'Hopital's Rule again we get
= lim_(x->0^(+)) ((e^x - 1)') / ((3x^2)')
= lim_(x->0^(+)) (e^x) / ((6x))
so now plugging the vale of x= 0 we get
= lim_(x->0^(+)) (e^x) / ((6x))
= (e^0) / ((6(0)))
= 1/0
= oo