#### Answer

$\dfrac{4}{3}$

#### Work Step by Step

We want to find $\lim\limits_{u \to 1}\dfrac{u^4-1}{u^3-1}$, but we can't use the quotient rule for limits because the limit of the denominator as $u$ approaches 1 is zero, and if we try to substitute $u=1$ directly, we get zero in both the numerator and the denominator.
But note that $\dfrac{u^4-1}{u^3-1}=\dfrac{(u-1)(u^3+u^2+u+1)}{(u-1)(u^2+u+1)}=\dfrac{(u^3+u^2+u+1)}{u^2+u+1}$
for all $u \neq 1.$
Thus $\lim\limits_{u \to 1}\dfrac{u^4-1}{u^3-1}=\lim\limits_{u \to 1}\dfrac{u^3+u^2+u+1}{u^2+u+1}=\dfrac{1+1+1+1}{1+1+1}=\dfrac{4}{3}.$