1. **State the problem:** Evaluate the limit $$\lim_{x \to 4} \frac{x^{2} - 16}{x^{3} - 64}$$. 2. **Recognize the form:** Substitute $x=4$ directly: $$\frac{4^{2} - 16}{4^{3} - 64} = \frac{16 - 16}{64 - 64} = \frac{0}{0}$$ which is an indeterminate form. So, we need to simplify. 3. **Factor numerator and denominator:** - Numerator: $x^{2} - 16 = (x - 4)(x + 4)$ (difference of squares). - Denominator: $x^{3} - 64 = (x - 4)(x^{2} + 4x + 16)$ (difference of cubes). 4. **Simplify the fraction:** $$\frac{(x - 4)(x + 4)}{(x - 4)(x^{2} + 4x + 16)}$$ Cancel the 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 of the simplified expression:** Substitute $x = 4$: $$\frac{4 + 4}{4^{2} + 4 \times 4 + 16} = \frac{8}{16 + 16 + 16} = \frac{8}{48} = \frac{1}{6}$$ **Final answer:** $$\lim_{x \to 4} \frac{x^{2} - 16}{x^{3} - 64} = \frac{1}{6}$$