Answer
$$\lim\limits_{u\to1}\frac{u^4-1}{u^3+5u^2-6u}=\frac{4}{7}$$
Work Step by Step
$$A=\lim\limits_{u\to1}\frac{u^4-1}{u^3+5u^2-6u}$$
- Consider the numerator:
$u^4-1=(u^2-1)(u^2+1)=(u-1)(u+1)(u^2+1)$
- Consider the denominator:
$u^3+5u^2-6u=u(u^2+5u-6)=u(u-1)(u+6)$
Therefore, $$A=\lim\limits_{u\to1}\frac{(u-1)(u+1)(u^2+1)}{u(u-1)(u+6)}$$$$A=\lim\limits_{u\to1}\frac{(u+1)(u^2+1)}{u(u+6)}$$$$A=\frac{(1+1)(1^2+1)}{1\times(1+6)}$$$$A=\frac{4}{7}$$
