1. **State the problem:** Find the limit $$\lim_{x\to 4} \frac{x^3 - 4x^2}{x^2 - 16}$$. 2. **Recall the formula and rules:** To find limits involving rational functions where direct substitution leads to an indeterminate form like $$\frac{0}{0}$$, we factor and simplify the expression. 3. **Evaluate the function at $$x=4$$:** $$\frac{4^3 - 4 \cdot 4^2}{4^2 - 16} = \frac{64 - 64}{16 - 16} = \frac{0}{0}$$ which is indeterminate. 4. **Factor numerator and denominator:** Numerator: $$x^3 - 4x^2 = x^2(x - 4)$$ Denominator: $$x^2 - 16 = (x - 4)(x + 4)$$ 5. **Simplify the expression by canceling common factors:** $$\frac{x^2 \cancel{(x - 4)}}{\cancel{(x - 4)} (x + 4)} = \frac{x^2}{x + 4}$$ 6. **Evaluate the simplified expression at $$x=4$$:** $$\frac{4^2}{4 + 4} = \frac{16}{8} = 2$$ 7. **Conclusion:** The limit is $$2$$. **Final answer:** 2