Answer
$$\lim_{t\to1}\frac{t^8-1}{t^5-1}=\frac{8}{5}$$
Work Step by Step
$$A=\lim_{t\to1}\frac{t^8-1}{t^5-1}$$
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:
$$A=\lim_{t\to1}\frac{(t^8-1)'}{(t^5-1)'}$$
$$A=\lim_{t\to1}\frac{8t^7}{5t^4}$$
$$A=\frac{8\times1^7}{5\times1^4}$$
$$A=\frac{8}{5}$$
