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Trouble with Limits

  1. Jan 16, 2007 #1
    (X^4-15X^3+32X+372X-1440)/X^2-X-30 as X approaches 6

    I know that somehow I am supposed to be able to factor this, but I'm having trouble doing so, and am stuck with a TI-82 that isn't very much help, either. Could someone please show me how to do this by hand?
     
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  3. Jan 16, 2007 #2

    marcusl

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    EDIT: Sorry, I was blind to the answer. mathwonk has it right in the next post...
     
    Last edited: Jan 16, 2007
  4. Jan 16, 2007 #3

    mathwonk

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    the root factor theorem says that if x=6 makes a polynomial zero, then x-6 is a factor, and vice versa.
     
  5. Jan 17, 2007 #4
    did you write the question correctly?
    is the third term in the numerator [itex]32x^2[/itex]?
    then it becomes
    [tex]\frac{x^4-15x^3+32x^2+372x-1440}{x^2-x-30}=\frac{(x+5)(x-8)(x-6)^2}{(x+5)(x-6)}=(x-8)(x-6)[/tex]

    and this approaches 0 as x approaches 6.
     
  6. Jan 17, 2007 #5

    Gib Z

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    I would have a feeling you assumption is correct, murshid. Nice work
     
  7. Jan 18, 2007 #6
    Thank you so much for your help!! All of you! It is very much apprecitated!
     
  8. Jan 18, 2007 #7
    Since the trouble was originally that you got 0/0 by plugging 6 in directly, you could also have used L'Hopital's rule. Of course, it is conceptually more elementary to factor the polynomial, but it may actually be less work to do 1 derivative than to do the polynomial long division.

    Just offering an alternative.

    --Stuart Anderson
     
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