1. **State the problem:** We need to find the limit \( \lim_{x \to -2} \frac{8 + x^3}{4 - x^2} \). 2. **Recall the limit evaluation method:** If direct substitution leads to a defined value, substitute \(x = -2\) directly. If it leads to an indeterminate form like \(\frac{0}{0}\), factor and simplify. 3. **Direct substitution:** Substitute \(x = -2\): $$\frac{8 + (-2)^3}{4 - (-2)^2} = \frac{8 - 8}{4 - 4} = \frac{0}{0}$$ which is indeterminate. 4. **Factor numerator and denominator:** - Numerator: \(8 + x^3 = 8 + (-2)^3 = 8 + (-8) = 0\), so factor as sum of cubes: $$8 + x^3 = (2)^3 + x^3 = (x + 2)(x^2 - 2x + 4)$$ - Denominator: \(4 - x^2 = (2)^2 - x^2 = (2 - x)(2 + x)\) 5. **Rewrite the limit:** $$\lim_{x \to -2} \frac{(x + 2)(x^2 - 2x + 4)}{(2 - x)(2 + x)}$$ Note that \(2 + x = x + 2\), so denominator is \((2 - x)(x + 2)\). 6. **Cancel common factor \(x + 2\):** $$\lim_{x \to -2} \frac{(x + 2)(x^2 - 2x + 4)}{(2 - x)(x + 2)} = \lim_{x \to -2} \frac{x^2 - 2x + 4}{2 - x}$$ 7. **Evaluate the simplified limit by substitution:** $$\frac{(-2)^2 - 2(-2) + 4}{2 - (-2)} = \frac{4 + 4 + 4}{2 + 2} = \frac{12}{4} = 3$$ **Final answer:** $$\boxed{3}$$