1. **State the problem:** Find the limit $$\lim_{x \to 1} \frac{x^3 - 3x + 2}{x^3 - 4x + 3}$$. 2. **Check direct substitution:** Substitute $x=1$ into numerator and denominator: $$1^3 - 3(1) + 2 = 1 - 3 + 2 = 0$$ $$1^3 - 4(1) + 3 = 1 - 4 + 3 = 0$$ Both numerator and denominator equal zero, so we have an indeterminate form $\frac{0}{0}$. 3. **Factor numerator and denominator:** Numerator: $x^3 - 3x + 2$ Try to factor by finding roots: Test $x=1$: $1 - 3 + 2 = 0$, so $(x-1)$ is a factor. Divide numerator by $(x-1)$: $$\frac{x^3 - 3x + 2}{x-1} = x^2 + x - 2$$ Factor $x^2 + x - 2$: $$(x+2)(x-1)$$ So numerator factors as: $$(x-1)(x+2)(x-1) = (x-1)^2 (x+2)$$ Denominator: $x^3 - 4x + 3$ Test $x=1$: $1 - 4 + 3 = 0$, so $(x-1)$ is a factor. Divide denominator by $(x-1)$: $$\frac{x^3 - 4x + 3}{x-1} = x^2 + x - 3$$ Factor $x^2 + x - 3$: No simple integer roots, so keep as is. Denominator factors as: $$(x-1)(x^2 + x - 3)$$ 4. **Rewrite the limit:** $$\lim_{x \to 1} \frac{(x-1)^2 (x+2)}{(x-1)(x^2 + x - 3)}$$ 5. **Cancel common factor $(x-1)$:** $$\lim_{x \to 1} \frac{\cancel{(x-1)} (x-1) (x+2)}{\cancel{(x-1)} (x^2 + x - 3)} = \lim_{x \to 1} \frac{(x-1)(x+2)}{x^2 + x - 3}$$ 6. **Evaluate the limit by direct substitution:** Substitute $x=1$: Numerator: $(1-1)(1+2) = 0 \times 3 = 0$ Denominator: $1^2 + 1 - 3 = 1 + 1 - 3 = -1$ 7. **Final answer:** $$\lim_{x \to 1} \frac{x^3 - 3x + 2}{x^3 - 4x + 3} = \frac{0}{-1} = 0$$