Answer
$\lim\limits_{v \to 2} \frac{v^{3}-8}{v^{4}-16}= \frac{3}{8} = 0.375$
Work Step by Step
Simplify:
$\frac{v^{3}-8}{v^{4}-16} = \frac{(v-2)(v^{2}+2v+4)}{(v+2)(v-2)(v^{2}+4)} = \frac{(v^{2}+2v+4)}{(v+2)(v^{2}+4)}$
Now,
$\lim\limits_{v \to 2} \frac{v^{3}-8}{v^{4}-16}= \lim\limits_{v \to 2} \frac{(v^{2}+2v+4)}{(v+2)(v^{2}+4)} = \frac{(2^{2} + 2(2)+ 4)}{(2+2)(2^{2}+4)} = \frac{12}{32} = \frac{3}{8} = 0.375$
