anonymous
  • anonymous
Evaluate the limit, if it exist lim as x approaches 1, (x^3-1)/(x^2-1).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
hint: exponents!
anonymous
  • anonymous
please show step by step thanks
amistre64
  • amistre64
3x/2x = 3/2

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anonymous
  • anonymous
i dont understand
amistre64
  • amistre64
it the general hospital rule...
anonymous
  • anonymous
ok factor top and bottom \[\frac{x^3-1}{x^2-1}=\frac{(x-1)(x^2+x+1)}{(x+1)(x-1)}\]
anonymous
  • anonymous
now cancel the x - 1 to get \[\frac{x^2+x+1}{x+1}\]
anonymous
  • anonymous
replace x by 1, get \[\frac{3}{2}\]
anonymous
  • 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
anonymous
  • anonymous
this kind of problem tends to come before differentiation and therefore before l'hopital's rule
anonymous
  • anonymous
difference of two cubes! \[a^3-b^3=(a-b)(a^2+ab+b^2)\]
anonymous
  • anonymous
in your case a = x and b = 1 to give \[x^3-1=(x-1)(x^2+x+1)\]
anonymous
  • anonymous
so basically you were factoring it out
anonymous
  • anonymous
exactly
anonymous
  • anonymous
ok cool i knew that what the steps were...well thanks again for the help
anonymous
  • 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.
anonymous
  • anonymous
yea.. thats what i did but somehow did it wrong
anonymous
  • 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
anonymous
  • anonymous
yea thats exactly what i did so i figure theres something more to it
anonymous
  • anonymous
thanks again your a life saver

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