## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{t\to1}\frac{t^4-1}{t^3-1}=\frac{4}{3}$
$A=\lim\limits_{t\to1}\frac{t^4-1}{t^3-1}$ Use $(a^2-b^2)=(a-b)(a+b)$ and $(a^3-b^3)=(a-b)(a^2+ab+b^2)$, we have $A=\lim\limits_{t\to1}\frac{(t^2-1)(t^2+1)}{(t-1)(t^2+t+1)}$ $A=\lim\limits_{t\to1}\frac{(t-1)(t+1)(t^2+1)}{(t-1)(t^2+t+1)}$ (again we use $(a^2-b^2)=(a-b)(a+b)$) $A=\lim\limits_{t\to1}\frac{(t+1)(t^2+1)}{t^2+t+1}$ (divide numerator and denominator by $t-1$) $A=\frac{(1+1)(1^2+1)}{1^2+1+1}$ $A=\frac{4}{3}$