1. **State the problem:** Find the limit $$\lim_{x \to 0} \frac{x^3 - 3x^2 + x}{x^3 - 2x}$$. 2. **Recall the limit rule:** If direct substitution leads to an indeterminate form like $$\frac{0}{0}$$, try to simplify the expression by factoring or algebraic manipulation. 3. **Check direct substitution:** Substitute $x=0$: $$\frac{0^3 - 3\cdot0^2 + 0}{0^3 - 2\cdot0} = \frac{0}{0}$$ 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. **Simplify the fraction:** $$\frac{x(x^2 - 3x + 1)}{x(x^2 - 2)} = \frac{x^2 - 3x + 1}{x^2 - 2}$$ for $x \neq 0$. 6. **Evaluate the simplified limit:** Substitute $x=0$: $$\frac{0^2 - 3\cdot0 + 1}{0^2 - 2} = \frac{1}{-2} = -\frac{1}{2}$$. 7. **Conclusion:** The limit is $$-\frac{1}{2}$$. This means as $x$ approaches 0, the value of the function approaches $$-\frac{1}{2}$$.