Calculus of a Single Variable, Chapter 8, 8.7, Section 8.7, Problem 33

Given to solve,
lim_(x->oo) lnx/(x^2)
as x->oo then the lnx/(x^2) =oo/oo 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->oo) lnx/(x^2)
= lim_(x->oo) ((lnx)')/((x^2)')
= lim_(x->oo) (1/x)/((2x))
= lim_(x->oo) (1/(2x^2))
upon plugging inx= oo , we get
= (1/(2(oo)^2))
= 1/oo = 0