Question

Evaluate the limx->3(2x^3-6x^2+x-3)/x-3

Answer 1

ok i know that it is indeterminant form...i.e. undefined at 3

Answer 2

if you replace x by 3 in the numerator, do you get 0? if so factor the numerator as
(x-3) something and cancel

Answer 3

yes i know to use algebra but i want to check to see if what i did is right

Answer 4

2x^2(x-3)+x-3/x-3

Answer 5

in fact numerators is
\[(x-3)(2x^2+1)\] so cancel and replace x by 3

Answer 6

i get 19

Answer 7

huh?

Answer 8

ok lets go slow

Answer 9

yes please...:) my algebra is rustic

Answer 10

numerator is
\[2x^3-6x^2+x-3\] yes?

Answer 11

yes

Answer 12

ok and now we also know that if you replace x by 3 you get 0

Answer 13

yes

Answer 14

"factor theorem" therefore tells you that the numerator is
\[2x^3-6x^2+x-3=(x-3)(something)\]

Answer 15

yes

Answer 16

your job is to find the "something"

Answer 17

because we have to cancel the denominator...

Answer 18

\(\underline{\text{limits are as limits does}}\)

Answer 19

and you had it when you wrote
\[2x^2(x-3)+x-3\]

Answer 20

okay...

Answer 21

you just didn't see it. both terms have a common factor of x - 3 so "factor it out" and get
\[(x-3)(2x^2+1)\]

Answer 22

your other option is to divide to see what you get, and your third option is just to thing what the second term must be, but you had it all along. now cancel the x-3 from top and bottom and replace x by 3 and be done

Answer 23

oh lord now guru is underlining. a little html is a dangerous thing.

Answer 24

i get it just needed to let it soak thanks so much satellite!!!!!!! and amistre

Answer 25

lol.

Answer 26

yw.

Answer 27

yup 19...

Answer 28

oh and i saw what i forgot 2x^2(x-3)+1(x-3)...lol

Answer 29

right. you had it and missed it.

Answer 30

lol...

Answer 31

whatcha looking at amistre?

Answer 32

shhhhh ... im napping lo

Answer 33

lol...

Answer 34

i knew you were hiding...

Answer 35

lo? yeah, lo! lo and behold ...\(\underline{\text{i nap}}\)