## Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to1}\frac{x^a-1}{x^b-1}=\frac{a}{b}\hspace{2cm}(b\ne0)$$
$$A=\lim_{x\to1}\frac{x^a-1}{x^b-1}\hspace{2cm}(b\ne0)$$ 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: $$A=\lim_{x\to1}\frac{(x^a-1)'}{(x^b-1)'}$$ $$A=\lim_{x\to1}\frac{ax^{a-1}}{bx^{b-1}}$$ $$A=\frac{a\times1^{a-1}}{b\times1^{b-1}}$$ $$A=\frac{a\times1}{b\times1}$$ $$A=\frac{a}{b}$$