Question

The limit of (X^3-1)/(X^2-1) as X aproaches 1. Doesn't exist, right?

Answer 1

it does exist.

Answer 2

.... How? They don't aprouch the same point and I can't get it so the x=1...

Answer 3

can u factor x^2-1 ?

Answer 4

I did that. I also did the difference of 2 cubes on top. I canceled out the (x+1) on top and bottom. And I ended up with (x-1) in the denominator. and if I put 1 in for x. it will equal 0.

Answer 5

x+1 doesn't get cancelled, x-1 gets cancelled.....
there is no x+1 in numerator....

Answer 6

can't you just do L'hopital 's rule?

Answer 7

\(\huge(x^3-1)=(x-1)(x^2+x+1)\)

Answer 8

ohhhhh! I forgot I was doing the difference for 2 cubes on the top. So I accidently made the top (x--1) instead of (x-1)... Thank yOU!

Answer 9

so it does exist, what value did u get ?

Answer 10

1/2

Answer 11

nopes.

Answer 12

T_T ok. hold on/ let me try again.

Answer 13

eee. I forgot to square the negative 1! This time i got 3/2... Better?

Answer 14

much better :)
infact accurate.

Answer 15

Thanks Goku! haha

Answer 16

welcome :)