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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    hint: exponents!

  2. anonymous
    • 5 years ago
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    please show step by step thanks

  3. amistre64
    • 5 years ago
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    3x/2x = 3/2

  4. anonymous
    • 5 years ago
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    i dont understand

  5. amistre64
    • 5 years ago
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    it the general hospital rule...

  6. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    now cancel the x - 1 to get \[\frac{x^2+x+1}{x+1}\]

  8. anonymous
    • 5 years ago
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    replace x by 1, get \[\frac{3}{2}\]

  9. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    this kind of problem tends to come before differentiation and therefore before l'hopital's rule

  11. anonymous
    • 5 years ago
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    difference of two cubes! \[a^3-b^3=(a-b)(a^2+ab+b^2)\]

  12. anonymous
    • 5 years ago
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    in your case a = x and b = 1 to give \[x^3-1=(x-1)(x^2+x+1)\]

  13. anonymous
    • 5 years ago
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    so basically you were factoring it out

  14. anonymous
    • 5 years ago
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    exactly

  15. anonymous
    • 5 years ago
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    ok cool i knew that what the steps were...well thanks again for the help

  16. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    yea.. thats what i did but somehow did it wrong

  18. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    yea thats exactly what i did so i figure theres something more to it

  20. anonymous
    • 5 years ago
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    thanks again your a life saver

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