1. **State the problem:** We need to find the limit $$\lim_{x \to 3} \frac{x^3 - 9x}{x^2 - 3x}$$. 2. **Analyze the expression:** 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 an indeterminate form. So, we need to simplify the expression. 3. **Factor numerator and denominator:** - Numerator: $$x^3 - 9x = x(x^2 - 9) = x(x - 3)(x + 3)$$ - Denominator: $$x^2 - 3x = x(x - 3)$$ 4. **Simplify the fraction by canceling common factors:** $$\frac{x(x - 3)(x + 3)}{x(x - 3)} = \frac{\cancel{x}\,\cancel{(x - 3)}(x + 3)}{\cancel{x}\,\cancel{(x - 3)}} = x + 3$$ 5. **Evaluate the simplified expression at $x=3$:** $$3 + 3 = 6$$ **Final answer:** $$\lim_{x \to 3} \frac{x^3 - 9x}{x^2 - 3x} = 6$$