Question

as the limit t approaches -2 at (t^3+8)/(t+2)

Answer 1

factor the top and it looks like you might be able to cancel t+2 in the denominator

Answer 2

is it 0?

Answer 3

you can do it throw a chart if you want to do it that way

Answer 4

is the limit 0?

Answer 5

Just do polynomial division! t+2 is a factor of t^3+8 because -2 is a zero of t^3+8

Answer 6

And then you just plug in t=-2 in the quadratic polynomial you get.

Answer 7

I did, i got 0.

Answer 8

12

Answer 9

is the answer.

Answer 10

\[\lim_{t \rightarrow -2} t^2-2t+4\]

Answer 11

btw correct term would be long division =P lol

Answer 12

(x^3 +y^3)=(x+y) (x^2 -xy + y^2)
t^3+8 is t^3 +2^3
(t+2)(t^2-2t+4) /(t+2) = (t^2-2t+4) provided t not equal to -2
as t gets close to -2 but is not equal to -2, the expression gets closer to 12.

Answer 13

if you want to be real smart call it polynomial long division if you have to get it right -.-