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Help on precalculus hw question

  1. Aug 29, 2016 #1
    1. The problem statement, all variables and given/known data
    limit as x approaches 2 of [ (x^2-cx+d)/(x^2-4) ] = 3. What is the value of c and d?

    2. Relevant equations
    Limits

    3. The attempt at a solution
    I tried cross multiplying to get (x^2 - cx + d) = 3x^2 - 12, and I was thinking of moving combining like terms and maybe factoring but I can't ever seem to get the right answer.
     
  2. jcsd
  3. Aug 29, 2016 #2

    Mark44

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    The denominator factors into (x - 2)(x + 2). If you assume that the numerator also has a factor of either x + 2 or x - 2, what would the other factor have to be to result in a limit of 3? An educated guess or two will be helpful.
     
  4. Aug 29, 2016 #3
    @Mark44
    Just for curiosity, can we apply L' hospital's rule here, assuming the numerator to be 0 when x→2 ?
     
  5. Aug 29, 2016 #4

    Mark44

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    I didn't think of doing that, but it seems to work. That technique wouldn't be available in a precalculus setting, which is what I assumed by where you the OP posted the question.
     
    Last edited: Aug 29, 2016
  6. Aug 29, 2016 #5

    SammyS

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    (@Molar isn't OP.)
     
  7. Aug 29, 2016 #6
    I tried that but I want to know how to solve it algebraically.
     
  8. Aug 29, 2016 #7

    SteamKing

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    I don't think that's possible.

    In the limit, the denominator ##x^2 - 4## goes to 0 as x → 2.

    Cross multiplying means that you are multiplying the RHS by zero as well, leaving ##x^2 - cx + d = 0##, for which there can be an infinite number o' solutions.

    Evaluating limit expressions sometimes takes subtlety, where applying algebra will fail or mislead you.
     
  9. Aug 29, 2016 #8
    Thanks! I ended up getting c = 8 and d = -20
     
  10. Aug 29, 2016 #9
    Thanks for your help!
     
  11. Aug 29, 2016 #10

    SammyS

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    I get a different sign for one of those.
     
  12. Aug 30, 2016 #11

    Ray Vickson

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    It is easiest to substitute ##x = 2 + h## in both the numerator and denominator. Your ratio must make sense as ##h \to 0##, and the limit must = 3. That gives you two conditions involving the two parameters ##c, d##.
     
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