Question

Limitquestion

Answer 1

So as x -> -3

Answer 2

\[\frac{ x^2-9 }{ x^3 + 3}\]

Answer 3

I know I can factor the top a little like (x-3)(x+3)

Answer 4

you can factor the bottom as well

Answer 5

sum of cubes

Answer 6

or if anything

you now x-3 is a factor, so divide it out of it

Answer 7

err, x+3 that is :)

Answer 8

Would the sum of cubes look like this?
(x + 3)(x^2 - cuberoot(3)x + squareroot(3) ) ?

Answer 9

did you try and plug in the number first?

Answer 10

-3x

Answer 11

(a+b)(a^2+b^2-ab)

a(a^2+b^2-ab) +b(a^2+b^2-ab)

a^3+abb -aab +aab+b^3-abb

a^3 -aab +aab+b^3 etc ...

Answer 12

So purple math describes sum of cubes as
"a^3 + b^3 = (a + b)(a^2 – ab + b^2)
a^3 – b^3 = (a – b)(a^2 + ab + b^2)"
a = x
b = 3 or cuberoot(3) ?

Answer 13

ima just wolf it to avoid my typos

Answer 14

or just plug in the number and realize you get 0 and avoid that
which is the answer

Answer 15

x^3 + 3

Answer 16

Yeah trying to do it algebraically though @Austin6i6

Answer 17

Instead of direct subsitution

Answer 18

-27+9 aint zero, so yeah

Answer 19

whats the numerator @amistre64

Answer 20

^^^

Answer 21

0

Answer 22

i was thinking the bottom was zero for some reason lol

Answer 23

whats 0/ any number @amistre64

Answer 24

0

Answer 25

Aiight algebraically is too much

Answer 26

hey, if they wanted me to pay attention in elementary school, they wouldnt have had all the girls around :)

Answer 27

Im just going to direct subsitution

Answer 28

problem is done, i guess if you want to practice, but its really pointless because thats how all mathematicians will do it.

Answer 29

it algebraic enough,

Answer 30

Ye

Answer 31

if (x+3) was a factor on the bottom, and it aint

we get: (x^3+9)= (x+3)(x^2 -3x +9) -18

Answer 32

if the remainder was 0, not -18 then we would have had an issue