Question

Evaluate the limit, if it exist
lim as x approaches 1, (x^3-1)/(x^2-1).

Answer 1

hint: exponents!

Answer 2

please show step by step thanks

Answer 3

3x/2x = 3/2

Answer 4

i dont understand

Answer 5

it the general hospital rule...

Answer 6

ok factor top and bottom
\[\frac{x^3-1}{x^2-1}=\frac{(x-1)(x^2+x+1)}{(x+1)(x-1)}\]

Answer 7

now cancel the x - 1 to get
\[\frac{x^2+x+1}{x+1}\]

Answer 8

replace x by 1, get \[\frac{3}{2}\]

Answer 9

oooooh ok i tried the other way and thought that you could factor it out...im still trying to understand where the x^2+x +1 came from

Answer 10

this kind of problem tends to come before differentiation and therefore before l'hopital's rule

Answer 11

difference of two cubes!
\[a^3-b^3=(a-b)(a^2+ab+b^2)\]

Answer 12

in your case a = x and b = 1 to give
\[x^3-1=(x-1)(x^2+x+1)\]

Answer 13

so basically you were factoring it out

Answer 14

exactly

Answer 15

ok cool i knew that what the steps were...well thanks again for the help

Answer 16

btw if you replace x by 1 in the numerator you get 0 yes? this means the numerator MUST factor as (x-1) times something.

Answer 17

yea.. thats what i did but somehow did it wrong

Answer 18

it is not a miracle you can factor, it is a consequence of the fact that you get 0 if you replace x by 1. same as in the denominator

Answer 19

yea thats exactly what i did so i figure theres something more to it

Answer 20

thanks again your a life saver