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Does a limit exist?

  1. Jul 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Is there a number 'a' such that

    (3x^2 + ax + a + 3) / (x^2 + x - 2)

    exists as x goes to -2? If so, find the value(s) of 'a' and the limit.

    2. Relevant equations
    The limit rules. Algebra.

    3. The attempt at a solution
    Well... I have absolutely no clue what I'm supposed to do. I suppose I'll have to choose my 'a' values in such a way so the denominator no longer equals zero, *but* other than brute-forcing a-integers in a graphical calculator I have little idea on how to do that.

    I began by factoring the denominator to (x+2)(x-1), with the hope that I may be able to negate the x+2 somehow, but that didn't work. Then I tried to get (x^2+x-2) in the numerator so the numerator and denominator would cancel out and leave a bit on top, but I couldn't find a way to do that, either.

    Could someone give me a clue on what approach to try here?
     
  2. jcsd
  3. Jul 6, 2012 #2

    micromass

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    The denominator always goes to 0 if x goes to -2. What should the numerator go to if the limit were defined?? For example, if the numerator were 1, then the limit would be "1/0" which would be an infinity and thus the limit would not exist.

    Is there a value for the numerator for which the limit does exist?
     
  4. Jul 6, 2012 #3
    Have you heard of L'Hopital's Rule? It should work here. If not, note that the numerator must also go to 0 as x goes to -2, and must therefore have a factor of x+2.
     
  5. Jul 6, 2012 #4
    Are you saying that the numerator should equal 0?
     
  6. Jul 6, 2012 #5

    micromass

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    Yes! If x=-2, then the numerator should be 0.
     
  7. Jul 6, 2012 #6
    Ok, let me see if I got this right:

    3(-2)^2+a(-2)+a+3 = 0
    -> 12 - 2a + a + 3 = 0
    -> a = 15

    Therefore (substituting a):
    lim[x->-2] (3x^2 + 15x + 18) / (x^2 + x - 2)
    = lim[x->-2] (3(x+2)(x+3)) / ((x+2)(x-1))
    = lim[x->-2] (3(x+3)) / (x-1)
    = 3 / -3
    = -1

    I think I got it. Thanks again! :smile:
     
  8. Jul 6, 2012 #7

    SammyS

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    That looks good.
     
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