Answer
$$ \lim _{x \rightarrow-2} \frac{x^{3}+8}{x^{2}+6 x+8}=6$$
Work Step by Step
Given $$ \lim _{x \rightarrow-2} \frac{x^{3}+8}{x^{2}+6 x+8}$$
let $$ f(x) = \frac{x^{3}+8}{x^{2}+6 x+8}$$
Since, we have
$$ f( -2)= \frac{-8+8}{4-12+8}=\frac{0}{0}$$
So, transform algebraically and cancel, we get
\begin{aligned}
L&= \lim _{x \rightarrow-2} \frac{x^{3}+8}{x^{2}+6 x+8}\\
&=\lim _{x \rightarrow-2} \frac{(x+2)\left(x^{2}-2 x+4\right)}{(x+2)(x+4)}\\
&=\lim _{x \rightarrow-2} \frac{\left(x^{2}-2 x+4\right)}{x+4}\\
&=\frac{4+4+4}{-2+4}\\
&=\frac{12}{2}\\
&=6\\
\end{aligned}
