Calculus
posted by John .
Is there a number "a" such that the limit as x approaches 2, f(x) approaches (3x^2+ax+a+3)/(x^2+x2) exist?

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. 
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)(x1))
x > 2
= lim 3(x+3)/(x1)
x > 2
= 3(1)/(3) = 1 
Thanks Reiny, I took the derivative a second time without checking if both numerator and denominator are zero.