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Find the value of the limit

  1. Aug 5, 2008 #1
    1. The problem statement, all variables and given/known data

    Is there a number a such that

    [tex]\lim_{x \rightarrow -2} \frac{3x^2+ax+a+3}{x^2+x-2}[/tex]

    exists? If so find the value of a and the value of the limit.


    I can't figure out how to find the value of a. I can guess and check but that will take forever. Is there some other way of doing this problem other then educated guesses and luck?
     
  2. jcsd
  3. Aug 5, 2008 #2
    Re: Limits

    There's a couple of things you can try with these kinds of cases. You know as x----> -2, the denominator goes to 0, so for the limit to exist, you either need to factor out an (x+2) from the denominator and cancel it out with an x+2 from the numerator(find an a that will allow you to do it), or you can alternatively approach it by finding an a that will make the limit in 0/0 form so you can apply L'hopital's rule
     
  4. Aug 5, 2008 #3
    Re: Limits

    I would use the quadratic formula to find the value of 'a' when one of the roots is -2. That way you can have a 0/0 limit as stated above.
     
  5. Aug 5, 2008 #4
    Re: Limits

    This is prior to L'hopital's rule, so there must be a way to factor it out like you said. Thanks for the help.
     
  6. Aug 5, 2008 #5
    Re: Limits

    a = 3, b = a, c = (a+3).

    Substitute these values into the quadratic formula and let it equal to -2 (so we could have a 0/0 limit). This will give you a value for a for which the limit will exist as x approaches -2. (I tested it and it works).
     
  7. Aug 5, 2008 #6
    Re: Limits

    Intuitively we want to cancel out the (x+2) factor from top and bottom. So it comes down to finding an (x+2) factor in the numerator. Now just use the fact that if (x-a) is a factor of polynomial f(x), then f(a) = 0.
     
  8. Aug 5, 2008 #7
    Re: Limits

    Not sure I quite follow you there.
     
  9. Aug 5, 2008 #8
    Re: Limits

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

    So I just need to focus on getting a factor of x+2 in the numerator. Is there a way in which I can do this other then just guessing numbers for a. Is there a way I can set this up to solve for a?
     
  10. Aug 5, 2008 #9

    rock.freak667

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    Re: Limits


    you want (x+2) to be a factor, use remainder and factor theorem.

    If [itex]f(x)=3x^2+2ax+a+3[/itex], what would f(-2) be?
     
  11. Aug 5, 2008 #10
    Re: Limits

    15-3a

    I don't know those theorems I will have to look them up.
     
  12. Aug 5, 2008 #11

    rock.freak667

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    Re: Limits

    If x+2 is a factor, then f(-2)=0.
     
  13. Aug 5, 2008 #12
    Re: Limits

    if I set that equal to 0, then a would equal 5. But that isn't the right answer.
     
  14. Aug 5, 2008 #13

    rock.freak667

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    Re: Limits

    Recheck this 3(-2)^2+(-2)a+a+3
     
  15. Aug 5, 2008 #14
    Re: Limits


    Oh sorry I was looking at that one. This isn't quite clicking with me. Yes that answer gave me the answer, but why? We said if x + 2 is a factor then f(-2) = 0, and we found a by setting f(-2) = 0, but we never showed that x + 2 is indeed a factor. So how are we sure that it is right?
     
  16. Aug 5, 2008 #15
    Re: Limits

    Well as rockfreak mentioned, it's technically called remainder/factor theorem. But really almost by definition (roots of polynomial equations, where function meets x-axis) if we have a factor (x+2) in a polynomial f(x), then it's clear that f(-2) = 0.

    As for the intuitive part, I'm sure you've encountered limits where if you try to substitute the value you get 0/0. Without knowing l'hopital's rule, the usual way is to try to cancel the same factor from the numerator and denominator (a removable discontinuity or graphically a "hole"). This is basically going to the other way around. We factor the denominator as (x+2)(x-1) and note that since it's the limit as x-> -2, it's the (x+2) factor that's the problem (a "hole" at x = -2). So we assume that there is an (x+2) factor in the numerator as well, which would imply that the expression in the numerator evaluated at x = -2 equals 0. Then we just solve for a.

    To check that it's right, you can factor the numerator (try it and see) and you'll see that everything works out and you'll find your limit.
     
  17. Aug 5, 2008 #16
    Re: Limits

    Instead of trying to factor the numerator and determining some value for "a", would it be better to say that if a solution exists for some value "a", then the numerator must go to zero? I say this because the denominator goes to zero anyways and this is the only condition that can lead to a solution. Therefore, apply L'Hopital and determine "a".

    EDIT: Let me rephrase this a little bit. If you find a factor to cancel with the numerator, then I suppose that method will work, but it seems like a lot of work. This statement seems better than my previous one: since the numerator goes to zero, "let" the numerator go to zero; therefore, we have 0/0 and apply L'Hopital.
     
    Last edited: Aug 5, 2008
  18. Aug 5, 2008 #17
    Re: Limits

    Well this is before L'hopital, so I guess they wanted it done the hard way.
     
  19. Aug 5, 2008 #18
    Re: Limits

    Well for one, this is pre-l'hopital's rule. Otherwise that would be an okay approach. But the idea is, you don't actually have to find a ridiculous factor. It really is the same approach as you mentioned in that we are supposing a solution exists. Denote p(x) as the numerator. ASSUMING that a solution exists, we have p(-2) = 0 SUPPOSING that p(x) does have an (x+2) factor. Then we just solve p(-2) = 0 => 3(-2)^2 -2a + a + 3 = 0 => a = 15.

    So you see we didn't have to go searching for factors, we only need to make sure our value of a works.
     
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