Answer
The limit exists when $a = 15$
$\lim\limits_{x \to -2}\frac{3x^2+15x+15+3}{x^2+x-2} = -1$
Work Step by Step
$\lim\limits_{x \to -2}\frac{3x^2+ax+a+3}{x^2+x-2}$
$=\lim\limits_{x \to -2}\frac{3x^2+ax+a+3}{(x+2)(x-1)}$
In this form, we can not evaluate this limit because there is $(x+2)$ in the denominator. If we could factor the numerator to include the factor $(x+2)$, then we could evaluate the limit.
Suppose the numerator has this form: $(3x+b)(x+2)$
Then:
$3x^2+ax+a+3 = (3x+b)(x+2)$
$3x^2+ax+a+3 = 3x^2+(b+6)x+2b$
$b+6 = a$
$b = a-6$
Then:
$2b = a+3$
$2(a-6) = a+3$
$a = 15$
Then:
$\lim\limits_{x \to -2}\frac{3x^2+ax+a+3}{x^2+x-2}$
$=\lim\limits_{x \to -2}\frac{3x^2+15x+15+3}{(x+2)(x-1)}$
$=\lim\limits_{x \to -2}\frac{3x^2+15x+18}{(x+2)(x-1)}$
$=\lim\limits_{x \to -2}\frac{(3x+9)(x+2)}{(x+2)(x-1)}$
$=\lim\limits_{x \to -2}\frac{3x+9}{x-1}$
$=\frac{3(-2)+9}{(-2)-1}$
$=\frac{3}{-3}$
$ = -1$
