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`