Answer
$$\lim_{x\to1}\frac{x^a-1}{x^b-1}=\frac{a}{b}\hspace{2cm}(b\ne0)$$
Work Step by Step
$$A=\lim_{x\to1}\frac{x^a-1}{x^b-1}\hspace{2cm}(b\ne0)$$
We know that $\lim_{x\to1}(x^a-1)=1^a-1=1-1=0$, and also $\lim_{x\to1}(x^b-1)=1^b-1=1-1=0$.
Therefore, this is an indeterminate form of $\frac{0}{0}$, applicable to the use of L'Hospital's Rule:
$$A=\lim_{x\to1}\frac{(x^a-1)'}{(x^b-1)'}$$
$$A=\lim_{x\to1}\frac{ax^{a-1}}{bx^{b-1}}$$
$$A=\frac{a\times1^{a-1}}{b\times1^{b-1}}$$
$$A=\frac{a\times1}{b\times1}$$
$$A=\frac{a}{b}$$
