Chapter 1 - Functions and Limits - 1.6 Calculating Limits Using the Limit Laws - 1.6 Exercises - Page 70: 19

$\frac{1}{12}$

$\lim\limits_{x \to -2}\frac{x+2}{x^3+8}=\frac{(-2)+2}{(-2)^3+8}=\frac{0}{0}=$indet so we have to do algebra $\lim\limits_{x \to -2}\frac{x+2}{x^3+8}=\lim\limits_{x \to -2}\frac{x+2}{(x+2)(x^2-2x+4)}=\lim\limits_{x \to -2}\frac{1}{x^2-2x+4}=\frac{1}{(-2)^2-2(-2)+4}=\frac{1}{12}$