Answer
$\frac{4}{3}$
Work Step by Step
$\lim\limits_{t \to 1}\frac{t^4-1}{t^3-1}=\frac{(1)^4-1}{(1)^3-1}=\frac{0}{0}=$indet
so we have to do algebra
$\lim\limits_{t \to 1}\frac{t^4-1}{t^3-1}=\lim\limits_{t \to 1}\frac{(t^2-1)(t^2+1)}{(t-1)(t^2+t+1)}=\lim\limits_{t \to 1}\frac{(t+1)(t-1)(t^2+1)}{(t-1)(t^2+t+1)}=\lim\limits_{t \to 1}\frac{(t+1)(t^2+1)}{t^2+t+1}=\frac{((1)+1)((1)^2+1)}{(1)^2+(1)+1}=\frac{4}{3}$
