Answer
-12
Work Step by Step
1. f is a closed function (we know by Th.10.1 that it is continuous), that is, L= $\displaystyle \lim_{x\rightarrow a}f(x)$ = $f(a)$,
for all a from the domain of f.
2. evaluating: $f(-2)$, (plugging $x=-2$) results in an indeterminate form 0/0.
So, we see that $x=-2$ is NOT in the domain of f.
We apply case 2 (of step 2). Simplifying $f(x)$,
$\displaystyle \frac{x^{3}+8}{x^{2}+3x+2}=\frac{x^{3}+2^{3}}{x^{2}+3x+2}=$... sum of cubes...
$=\displaystyle \frac{(x+2)(x^{2}-2x+4)}{x^{2}+3x+2}$ ... factor the denominator...
$=\displaystyle \frac{(x+2)(x^{2}-2x+4)}{(x+2)(x+1)} =$...reduce...$=\displaystyle \frac{x^{2}-2x+4}{x+1}=g(x)$
$g(-2)=\displaystyle \frac{(-2)^{2}-2(-2)+4}{(-2)+1}=\frac{12}{-1}=-12$
($-2$ is in the domain of g, g is closed-form), so
$\displaystyle \lim_{x\rightarrow-2}f(x)=\lim_{x\rightarrow-2}g(x)=g(-2)=-12$
