Answer
$\lim\limits_{u \to 1}\frac{u^{4}-1}{u^{3}-1} = \frac{4}{3} = 1.33$
Work Step by Step
Simplify:
$\frac{u^{4} - 1^{4}}{u^{3}-1} =\frac{ (u+1)(u-1)(u^{2}+1^{2})}{(u-1)(u^{2}+u+1)} = \frac{ (u+1)(u^{2}+1^{2})}{(u^{2}+u+1)}$
Now,
$\lim\limits_{u \to 1}\frac{u^{4}-1}{u^{3}-1} = \lim\limits_{u \to 1}\frac{ (u+1)(u^{2}+1^{2})}{(u^{2}+u+1)} = \frac{(1+1)(1+1)}{1+1+1} = \frac{4}{3} = 1.33$
