## Calculus 8th Edition

$\dfrac{8}{5}$
Given: $\lim\limits_{t\to1}\dfrac{t^8-1}{t^5-1}$ Here, $\lim_{t\to1}(t^8-1)=0$ and $\lim_{t\to1}(t^5-1)=1^5-1=0$, This shows an indeterminate form of type $\dfrac{0}{0}$, and so we will apply L'Hospital Rule. $\lim\limits_{t\to1}\dfrac{(t^8-1)'}{(t^5-1)'}=\lim\limits_{t\to1}\dfrac{8t^7}{5t^4}=\dfrac{8(1^7)}{5(1^4)}=\dfrac{8}{5}$