Answer
12
Work Step by Step
$\lim\limits_{h \to 0}\frac{(2+h)^3-8}{h}=\frac{(2+(0))^3-8}{(0)}=\frac{0}{0}=$indet
so we have to do algebra
$\lim\limits_{h \to 0}\frac{(2+h)^3-8}{h}=\lim\limits_{h \to 0}\frac{h^3+6h^2+12h+8-8}{h}=\lim\limits_{h \to 0}h^2+6h+12=(0)^2+6(0)+12=12$
