1. **State the problem:** Find the limit $$\lim_{x \to 0} \frac{x^3 - 3x^2 + x}{x^3 - 2x}$$. 2. **Recall the limit rules:** When direct substitution results in an indeterminate form like $$\frac{0}{0}$$, we try to simplify the expression by factoring or algebraic manipulation. 3. **Substitute $x=0$ directly:** $$\frac{0^3 - 3\cdot0^2 + 0}{0^3 - 2\cdot0} = \frac{0}{0}$$ which is indeterminate. 4. **Factor numerator and denominator:** - Numerator: $$x^3 - 3x^2 + x = x(x^2 - 3x + 1)$$ - Denominator: $$x^3 - 2x = x(x^2 - 2)$$ 5. **Rewrite the limit:** $$\lim_{x \to 0} \frac{x(x^2 - 3x + 1)}{x(x^2 - 2)}$$ 6. **Cancel common factor $x$ (valid since $x \to 0$, $x \neq 0$):** $$\lim_{x \to 0} \frac{x^2 - 3x + 1}{x^2 - 2}$$ 7. **Substitute $x=0$ now:** $$\frac{0^2 - 3\cdot0 + 1}{0^2 - 2} = \frac{1}{-2} = -\frac{1}{2}$$ **Final answer:** $$\boxed{-\frac{1}{2}}$$