Calculus of a Single Variable, Chapter 8, 8.7, Section 8.7, Problem 18

Given to solve,
lim_(x->1) (x^a - 1)/(x^b-1)
as x->1 then the lim_(x->1) (x^a - 1)/(x^b-1) =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) = 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->1) (x^a - 1)/(x^b-1)
= lim_(x->1) ((x^a - 1)')/((x^b-1)')
= lim_(x->1) (a(x^(a-1)))/((b(x^(b-1))))
now plugging the value of x = 1 then we get
= lim_(x->1) (ax^(a-1))/((bx^(b-1)))
= (a(1)^(a-1))/((b(1)^(b-1)))
= a/b