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OpenStudy (anonymous):

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

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OpenStudy (anonymous):

factor and cancel again

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

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OpenStudy (anonymous):

hold the phone

OpenStudy (anonymous):

???

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

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