## anonymous 5 years ago Evaluate the limit, if it exist lim as x approaches 1, (x^3-1)/(x^2-1).

1. anonymous

hint: exponents!

2. anonymous

please show step by step thanks

3. amistre64

3x/2x = 3/2

4. anonymous

i dont understand

5. amistre64

it the general hospital rule...

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

so basically you were factoring it out

14. anonymous

exactly

15. anonymous

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

16. anonymous

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.

17. anonymous

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

18. anonymous

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

19. anonymous

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

20. anonymous

thanks again your a life saver