Answer
\[a=15\]
and \[\lim_{x\rightarrow -2}\frac{3x^2+ax+a+3}{x^2+x-2}=-1\]
Work Step by Step
Let \[L=\lim_{x\rightarrow -2}\frac{3x^2+ax+a+3}{x^2+x-2}\]
Since \[\lim_{x\rightarrow -2}(x^2+x-2)=(-2)^2+(-2)-2=0\]
So denominator of $L$ tends to $0$ as $x\rightarrow -2$ so $L$ can exist only if
\[\lim_{x\rightarrow -2}(3x^2+ax+a+3)=0\]
\[\Rightarrow 3(-2)^2+a(-2)+a+3=12-2a+a+3=15-a=0\]
\[Rightarrow a=15\]
\[L=\lim_{x\rightarrow -2}\frac{3x^2+15x+15+3}{x^2+x-2}\]
\[L=\lim_{x\rightarrow -2}\frac{3(x^2+5x+6)}{x^2+2x-x-2}\]
\[L=\lim_{x\rightarrow -2}\frac{3(x^2+5x+6)}{(x+2)(x-1)}\]
\[L=\lim_{x\rightarrow -2}\frac{3(x+2)(x+3)}{(x+2)(x-1)}\]
\[L=\lim_{x\rightarrow -2}\frac{3(x+3)}{x-1}\]
\[L=\frac{3(-2+3)}{-2-1}=-1\]
