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LIMITS, not able to factor

  1. Sep 16, 2008 #1
    the question is:

    The limit as x approaches -2 when f(x)= x+2/x3+8

    I cannot factor it, or use a limit law (to my knowledge) and am simply having trouble finding the answer which is suppose to be 1/12.

    Any help getting me started would be greatly appreciated

    Susan
     
  2. jcsd
  3. Sep 16, 2008 #2

    morphism

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    Can you factor a^3 - b^3?
     
  4. Sep 16, 2008 #3

    statdad

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    Actually, you need to factor

    [tex]
    a^3 + b^3
    [/tex]

    instead of the difference of two cubes (assuming your function is

    [tex]
    f(x) = \frac{x+2}{x^3 + 8}
    [/tex]
    )
     
  5. Sep 16, 2008 #4

    morphism

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    It's the same thing, because a^3 + b^3 = a^3 - (-b)^3. :wink:

    The reason I chose to use difference instead of sum is because difference is usually more familiar, and might spark the right idea.
     
  6. Sep 16, 2008 #5

    statdad

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    Certainly true and obvious when you know the steps that are to be done. I've been teaching mathematics and statistics long enough to realize that occasionally a student may not make that connection - not because of lack of ability, but because of frustration, nerves, haste, or many other reasons. In a situation like this, a little more direct approach can't hurt.
    My previous post was not intended to be smarmy - I hope that no offense was taken, because none was meant.
     
  7. Sep 16, 2008 #6
    thank you so much both of you, I really had no idea what I could branch out of the equation and I would never have found that on my own
     
  8. Sep 17, 2008 #7

    statdad

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    You are welcome Susan_t. Good luck with your studies.
     
  9. Sep 17, 2008 #8

    HallsofIvy

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    WHENEVER a polynomial is 0 at x= a, then it has a factor of (x-a). Since x3+ 1 is 0 at x= -1, it MUST have (x+1) as a factor. Divide x3+ 1 by x+ 1 to find the other factor (which is, of course, x2+ x+ 1).

    Notice that says that any time you have a ratio of polynomial and both numerator and denominator are 0 at x= a, each MUST have a factor of x-a which you can then cancel.
     
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