Answer
L'Hospital's rule applies.
limit = $ -12$
Work Step by Step
Theorem 11.3 L'Hospital's Rule:
If $f$ and $g$ are differentiable functions such that
substituting $x=a$ in the expression $\displaystyle \frac{f(x)}{g(x)}$ gives
the indeterminate form $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$, then $\displaystyle \lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a} \displaystyle \frac{f^{\prime}(x)}{\mathrm{g}^{\prime}(x)}$.
That is, we can replace $f(x)$ and $g(x)$ with their derivatives and try again to take the limit.
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When x approaches $-2$, (evaluating for x=$-2$)
the numerator approaches $(-2)^{3}+8=-8+8=0$
the denominator approaches $4-6+2=0$
$\displaystyle \frac{f(x)}{g(x)}$ is of indeterminate form $\displaystyle \frac{0}{0}$,
L'Hospital's rule applies.
$\displaystyle \lim_{x\rightarrow-2}\frac{x^{3}+8}{x^{2}+3x+2}=\lim_{x\rightarrow-2} \displaystyle \frac{[x^{3}+8]^{\prime}}{[x^{2}+3x+2]^{\prime}}$
$=\displaystyle \lim_{x\rightarrow-2} \displaystyle \frac{3x^{2}+0}{2x+3+0}=$
$=\displaystyle \frac{3(2^{2})}{-4+3}=\frac{12}{-1}=-12$
