1. **State the problem:** Find the limit $$\lim_{x \to -2} \frac{x^2 - 4}{x^3 + 8}$$. 2. **Recall the formulas and rules:** - Factor the numerator and denominator if possible. - If direct substitution leads to an indeterminate form $\frac{0}{0}$, factor and simplify. 3. **Factor numerator and denominator:** - Numerator: $x^2 - 4 = (x - 2)(x + 2)$ (difference of squares). - Denominator: $x^3 + 8 = (x + 2)(x^2 - 2x + 4)$ (sum of cubes). 4. **Rewrite the limit:** $$\lim_{x \to -2} \frac{(x - 2)(x + 2)}{(x + 2)(x^2 - 2x + 4)}$$ 5. **Cancel common factors:** $$\lim_{x \to -2} \frac{(x - 2)\cancel{(x + 2)}}{\cancel{(x + 2)}(x^2 - 2x + 4)} = \lim_{x \to -2} \frac{x - 2}{x^2 - 2x + 4}$$ 6. **Substitute $x = -2$ into the simplified expression:** - Numerator: $-2 - 2 = -4$ - Denominator: $(-2)^2 - 2(-2) + 4 = 4 + 4 + 4 = 12$ 7. **Calculate the limit:** $$\frac{-4}{12} = -\frac{1}{3}$$ **Final answer:** $$\boxed{-\frac{1}{3}}$$