<s>[INST] finding the $\lim_{x \rightarrow -2 }\frac{x+2}{x^3+8}$ How do you solve $\lim_{x \rightarrow -2} \frac{x+2}{x^3+8}$ to get 1/12? I tried factoring out the denominator into $(x+2)(x+2)(x+2)$ and cancelling it out with the top but when you plug in 0 for x the denominator is still 0. 
 [/INST] Here are the steps
$$ \lim_{x\to -2} \frac{x+2}{x^3+8}=\lim_{x\to -2} \frac{x+2}{(x+2)(x^2-2x+4)} $$
$$ =\lim_{x\to -2} \frac{1}{x^2-2x+4}= \frac{1}{(-2)^2-2(-2)+4}=\frac{1}{4+4+4}=\frac{1}{12} $$
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