Calculus of a Single Variable, Chapter 8, 8.7, Section 8.7, Problem 34

Givne to solve ,
lim_(x->oo)ln(x^4)/x^3
= lim_(x->oo) 4 ln(x)/x^3
= 4lim_(x->oo) ln(x)/x^3
as x->oo then the ln(x)/x^3 =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
4lim_(x->oo) ln(x)/x^3
=4lim_(x->oo) (ln(x)')/((x^3)')
= 4lim_(x->oo) (1/x)/(3x^2)
= 4lim_(x->oo) (1/(3x^3))
so on plugging the value x= oo we get
= 4(1/(3(oo)^3))
= 0