Calculus
posted by Blue .
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.)

factor the bottom
(x3)(x2)
factor the top with (x3) a factor so we can cancel it
(x3)(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)
(x3)(x+4) / [(x3)(x2)]
(x+4)/(x2)
when x = 3
7/1 = 7 
hmmm. looks familiar. Anyway,
denominator is (x3)(x2), so we want the top to be
(x3)(xk) 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 = (x3)(x+4)