Answer
$$
\begin{aligned}
\lim _{x \rightarrow 1} \frac{x^{a}-1}{x^{b}-1} & = \quad[\text { for } b \neq 0]
\\
&=\frac{a}{b}
\end{aligned}
$$
Work Step by Step
$$
\lim _{x \rightarrow 1} \frac{x^{a}-1}{x^{b}-1} \quad , \quad b \ne 0
$$
Since
$$
\lim _{x \rightarrow 1} ( x^{a}-1)=0
$$
and
$$
\lim _{x \rightarrow 1} ( x^{b}-1)=0, \text{where } [ b \ne 0]
$$
the limit is an indeterminate form of type $\frac{0}{0}$ so we can apply l’Hospital’s Rule:
$$
\begin{aligned}
\lim _{x \rightarrow 1} \frac{x^{a}-1}{x^{b}-1} & \quad[\text { for } b \neq 0]
\\
& \stackrel{11}{=} \lim _{x \rightarrow 1} \frac{a x^{a-1}}{b x^{b-1}} \\
&=\frac{a(1)}{b(1)} \\
&=\frac{a}{b}
\end{aligned}
$$
