Answer
$$\frac{a}{b}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^a} - 1}}{{{x^b} - 1}} \cr
& {\text{Evaluating, we get:}} \cr
& = \frac{{{1^a} - 1}}{{{1^b} - 1}} = \frac{{1 - 1}}{{1 - 1}} = \frac{0}{0} \cr
& {\text{Using l'Hopital's rule, we get:}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to 1} \frac{d}{{dx}}\left( {{x^a} - 1} \right)}}{{\mathop {\lim }\limits_{x \to 1} \frac{d}{{dx}}\left( {{x^b} - 1} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 1} \frac{{a{x^{a - 1}}}}{{b{x^{b - 1}}}} \cr
& {\text{Evaluating the limit, we get:}} \cr
& = \frac{{a{{\left( 1 \right)}^{a - 1}}}}{{b{{\left( 1 \right)}^{b - 1}}}} \cr
& = \frac{a}{b} \cr
& {\text{Then}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^a} - 1}}{{{x^b} - 1}} = \frac{a}{b} \cr} $$
