Calculus of a Single Variable, Chapter 8, 8.7, Section 8.7, Problem 25

Given to solve,
lim_(x->oo) (x^2 +4x+7)/(x-6)
as x->oo then the (x^2 +4x+7)/(x-6) =oo/oo is of indeterminate form
so upon applying the L 'Hopital rule we get the solution as follows,
If 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) (x^2 +4x+7)/(x-6)
= lim_(x->oo) ((x^2 +4x+7)')/((x-6)')
= lim_(x->oo) (2x+4)/(1)
by plugging the value x=oo , we get
= 2(oo)+4
= oo