1. **State the problem:** Find the limit $$\lim_{x\to 3} \frac{x^3 - 9x}{x^2 - 3x}$$. 2. **Recall the formula and rules:** When direct substitution results in an indeterminate form like $$\frac{0}{0}$$, we simplify the expression by factoring and canceling common factors. 3. **Substitute $x=3$ directly:** $$\frac{3^3 - 9 \cdot 3}{3^2 - 3 \cdot 3} = \frac{27 - 27}{9 - 9} = \frac{0}{0}$$ which is indeterminate. 4. **Factor numerator and denominator:** $$x^3 - 9x = x(x^2 - 9) = x(x-3)(x+3)$$ $$x^2 - 3x = x(x-3)$$ 5. **Rewrite the limit expression:** $$\lim_{x\to 3} \frac{x(x-3)(x+3)}{x(x-3)}$$ 6. **Cancel common factors:** $$\lim_{x\to 3} \frac{\cancel{x}\cancel{(x-3)}(x+3)}{\cancel{x}\cancel{(x-3)}} = \lim_{x\to 3} (x+3)$$ 7. **Evaluate the simplified limit:** $$3 + 3 = 6$$ **Final answer:** $$\boxed{6}$$