1. **State the problem:** Evaluate the limit $$\lim_{x \to 4} \frac{x^2 - 16}{x^3 - 64}$$. 2. **Recognize the indeterminate form:** Substitute $x=4$ directly: $$\frac{4^2 - 16}{4^3 - 64} = \frac{16 - 16}{64 - 64} = \frac{0}{0}$$ which is indeterminate, so we need to simplify. 3. **Factor numerator and denominator:** - Numerator: $x^2 - 16 = (x-4)(x+4)$ - Denominator: $x^3 - 64 = (x-4)(x^2 + 4x + 16)$ (difference of cubes formula) 4. **Simplify the fraction by canceling common factor $(x-4)$:** $$\frac{\cancel{(x-4)}(x+4)}{\cancel{(x-4)}(x^2 + 4x + 16)} = \frac{x+4}{x^2 + 4x + 16}$$ 5. **Evaluate the limit by direct substitution:** $$\frac{4+4}{4^2 + 4 \times 4 + 16} = \frac{8}{16 + 16 + 16} = \frac{8}{48} = \frac{1}{6}$$ **Final answer:** $$\lim_{x \to 4} \frac{x^2 - 16}{x^3 - 64} = \frac{1}{6}$$