Calculus: Early Transcendentals 8th Edition

$$\lim_{t\to1}\frac{t^8-1}{t^5-1}=\frac{8}{5}$$
$$A=\lim_{t\to1}\frac{t^8-1}{t^5-1}$$ We see that $\lim_{t\to1}(t^8-1)=1^8-1=0$ and $\lim_{t\to1}(t^5-1)=1^5-1=0$, so this is an indeterminate form of $\frac{0}{0}$, which means we can apply L'Hospital's Rule: $$A=\lim_{t\to1}\frac{(t^8-1)'}{(t^5-1)'}$$ $$A=\lim_{t\to1}\frac{8t^7}{5t^4}$$ $$A=\frac{8\times1^7}{5\times1^4}$$ $$A=\frac{8}{5}$$