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Is there a number "a" such that the limit as x approaches -2, f(x) approaches (3x^2+ax+a+3)/(x^2+x-2) exist?

  • Calculus -

    As x → -2, the denominator approaches zero.
    Fo the limit to exist, the numerator must also be zero, which means:
    if f(x)=(3x²+ax+a+3)
    the f(-2)=0
    So solve for a in f(-2)=0, AND check using l'Hôpital's rule that the limit exists for the assumed value of a. You may need to apply l'Hôpital's rule more than once.

    I have 3 as the limit.

  • Calculus -

    yes, if a=15

    Since the limit exists, both numerator and denominator must have the factor x+2
    by synthetic division
    (3x^2+ax+a+3) รท (x+2) leaves no remainder, so a = 15
    then the question becomes
    lim (3x^2 + 15x + 18)/(x^2 + x - 2)
    x ---> -2
    = lim (3(x+2)(x+3)/((x+2)(x-1))
    x ---> -2
    = lim 3(x+3)/(x-1)
    x ---> -2
    = 3(1)/(-3) = -1

  • Calculus -

    Thanks Reiny, I took the derivative a second time without checking if both numerator and denominator are zero.

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