Answer:
Step-by-step explanation:
In order for the limit to exist at x=-2, the denominator factor of (x+2) must be canceled by a numerator factor of (x+2). Synthetic division (see below) shows the remainder when the numerator is divided by (x+2) is (15-a). In order for that remainder to be zero, so that (x+2) is a factor, we must have ...
[tex]15-a=0\\\\\boxed{a=15}[/tex]
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The same synthetic division shows the other numerator factor to be ...
(3x +(a -6))
For a=15, this is ...
(3x +9)
and the expression becomes ...
[tex]L= \lim\limits_{x \to -2} \dfrac{(3x+9)(x+2)}{(x-1)(x+2)}=\dfrac{3(-2)+9}{-2-1}=\dfrac{3}{-3}\\\\\boxed{L=-1}[/tex]
