Answer
$\dfrac {3}{16}$
Work Step by Step
$\lim _{v\rightarrow 2}\dfrac {v^{2}+2v-8}{v^{4}-16}=\dfrac {\left( v+4\right) \left( v-2\right) }{\left( v^{2}-4\right) \left( v^{2}+4\right) }=\dfrac {\left( v+4\right) \left( v-2\right) }{\left( v-2\right) \left( v+2\right) \left( v^{2}+4\right) }=\dfrac {\left( v+4\right) }{\left( v+2\right) \left( v^{2}+4\right) }=\dfrac {3}{16}$
