Question

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

Answer 1

factor and cancel again

Answer 2

\[\frac{(u-2)(u^2+2u+4)}{(u-2)(u+2)(u^2+4)}\]
\[\frac{u^2+2u+4}{(u+2)(u^2+4)}\]

Answer 3

replace u by 2 get
\[\frac{2^2+2\times 2+4}{(2+2)(2^2+4)}\]

Answer 4

i get
\[\frac{16}{32}=\frac{1}{2}\]

Answer 5

ooooo... now i get the idea... thx

Answer 6

hold the phone

Answer 7

???

Answer 8

if you replace u by 2 in top and bottom you get zero over zero. this means 2 is a root (zero) of both top and bottom. this means you can factor both as (x-2) times something and cancel the x-2 from top and bottom.

just thought i would mention it

Answer 9

all i am saying is that it always works that way. you got it