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Limits for Quadratic Equations

  1. Nov 7, 2008 #1
    1. The problem statement, all variables and given/known data



    2. Relevant equations
    [tex]
    \lim(\frac{(x^2)-2x-3}{(x^2)+5x+6})[/tex]
    [tex]
    x->(-2)
    [/tex]



    3. The attempt at a solution


    anyone can help out on this one?
     
  2. jcsd
  3. Nov 7, 2008 #2

    Dick

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    Factor the quadratics. Which factor is going to zero? Can you cancel it?
     
  4. Nov 7, 2008 #3
    er by factoring it i actually get

    [tex]

    \lim(\frac{(x-3)(x+1)}{(x+3)(x+2)})[/tex]
    [tex]

    x->(-2)
    [/tex]

    am stuck from here on~
     
  5. Nov 7, 2008 #4

    Dick

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    Ok, it doesn't cancel. My mistake. But the limit of the numerator is 5 and the limit of the denominator is 0. What does that tell you about the limit of the quotient?
     
  6. Nov 7, 2008 #5
    tends to infinity?
     
  7. Nov 7, 2008 #6

    Dick

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    It's worse than that. It goes to +infinity from one side and -infinity from the other. I would just say 'doesn't exist'.
     
  8. Nov 7, 2008 #7
    so there is a problem in the question? haha~
     
  9. Nov 7, 2008 #8

    Dick

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    Not necessarily. They can ask you to find a limit and expect you to determine that it doesn't exist. That's a possible correct answer.
     
  10. Nov 7, 2008 #9
    hmm icic~ troublesome question that got me stuck for so long haha~


    anyways, appreciate the help given, thanks so much~
     
  11. Nov 7, 2008 #10
    L'Hopitals?

    ( 2x - 2 ) / ( 2x + 5 ) -> -2 is -6

    Just realised L'Hop wont work cause it isn't 0/0. Sorry.

    or

    if i divide through by the highest power I get -5/12.

    I don't have anything to graph with, but plot it and see.
     
    Last edited: Nov 7, 2008
  12. Nov 7, 2008 #11

    Dick

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    l'Hopital, NO. And you know why. Dividing through by the highest power doesn't help. x doesn't go to infinity, you can't say the other terms go to zero. Plotting it will show there is no limit. As you should guess, the form is 5/0.
     
  13. Nov 8, 2008 #12

    HallsofIvy

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    It the numerator and denomenator separately go to 0, you can use L'Hopital. If the denominator goes to 0 and the numerator doesn't, the limit does not exist.
     
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