Question

lim (v^3-8)/(v^4-16)
v->2

Answer 1

\[\lim_{v\to2}\frac{v^3-8}{v^4-16}=\frac{0}{0}\]You can use L'Hôpital's rule\[\lim_{v\to2}\frac{3v^2}{4v^3}=\frac{12}{32}=\frac{3}{8}\]

Answer 2

Break down the top as a difference of cubes and the bottom as a difference of squares (v^2+4)(v^2-4), then break that v^2-4 down again into a difference of squares, and you should get the negative quantities to cancel, then just plug in 2.