Calculus

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Find the constant c so that
lim [x^2 + x + c]/[x^2 - 5x + 6] exists.
x->3

For that value of c, determine the limit. (Hint: Find the value of c for which x - 3 is a factor of the numerator.)

  • Calculus -

    factor the bottom
    (x-3)(x-2)

    factor the top with (x-3) a factor so we can cancel it
    (x-3)(x+b) = x^2+x+c
    x^2 - 3x +b x - 3 b = x^2 + x + c
    so
    b x -3 x = x so b = 4
    then c = -12
    so
    (x^2 + x - 12)/(x^2 - 5 x + 6)

    (x-3)(x+4) / [(x-3)(x-2)]

    (x+4)/(x-2)
    when x = 3
    7/1 = 7

  • Calculus -

    hmmm. looks familiar. Anyway,

    denominator is (x-3)(x-2), so we want the top to be
    (x-3)(x-k) so that the fraction exists everywhere except at x=3, but has a finite limit there.

    x^2 - (k+3)x + 3k = x^2 + x + c
    k+3 = -1
    k = -4

    so c = 3k = -12

    x^2 + x + c = x^2 + x -12 = (x-3)(x+4)

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