You need to substitute 1 for x in equation of function such that:

`lim_(x-gt1)(x^a - 1)/(x^b - 1) = (1-1)/(1-1) = 0/0`

Since the limit in indeterminate, you may use l'Hospital's theorem to evaluate the limit such that:

`lim_(x-gt1)(x^a - 1)/(x^b - 1) = lim_(x-gt1)((x^a - 1)')/((x^b - 1)')`

`lim_(x-gt1)((x^a - 1)')/((x^b - 1)') = lim_(x-gt1)(ax^(a-1))/(bx^(b-1))`

`lim_(x-gt1)(ax^(a-1))/(bx^(b-1)) = (a/b)lim_(x-gt1)(x^(a-1))/(x^(b-1))`

You need to substitute 1 for x in equation of function such that:

`(a/b)lim_(x-gt1)(x^(a-1))/(x^(b-1)) = (a/b)(1^(a-1))/(1^(b-1))` `(a/b)lim_(x-gt1)(x^(a-1))/(x^(b-1)) = (a/b)(1/1)`

`(a/b)lim_(x-gt1)(x^(a-1))/(x^(b-1)) = (a/b)`

**Hence, evaluating the limit of the function if x approches to 1 yields `lim_(x-gt1)(x^a - 1)/(x^b - 1) = (a/b).` **

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