1. **Problem:** Evaluate the limit $$\lim_{x \to 2} f(x)$$ where $$f(x) = \frac{x^3 - 4x^2 + x + 6}{x^3 - 6x^2 + 11x - 6}$$. 2. **Formula and rules:** To find the limit of a rational function as $$x$$ approaches a value, first try direct substitution. If it results in an indeterminate form like $$\frac{0}{0}$$, factor numerator and denominator and simplify. 3. **Direct substitution:** Substitute $$x=2$$: $$\text{Numerator} = 2^3 - 4(2)^2 + 2 + 6 = 8 - 16 + 2 + 6 = 0$$ $$\text{Denominator} = 2^3 - 6(2)^2 + 11(2) - 6 = 8 - 24 + 22 - 6 = 0$$ We get $$\frac{0}{0}$$, an indeterminate form. 4. **Factor numerator:** $$x^3 - 4x^2 + x + 6$$ Group terms: $$= (x^3 - 4x^2) + (x + 6) = x^2(x - 4) + 1(x + 6)$$ Try factoring by grouping or synthetic division. Testing $$x=2$$ as root: $$2^3 - 4(2)^2 + 2 + 6 = 0$$ confirms $$x-2$$ is a factor. Divide numerator by $$x-2$$: Using synthetic division: Coefficients: 1, -4, 1, 6 Bring down 1, multiply by 2: 2, add to -4: -2 Multiply by 2: -4, add to 1: -3 Multiply by 2: -6, add to 6: 0 Quotient: $$x^2 - 2x - 3$$ So numerator factors as: $$ (x - 2)(x^2 - 2x - 3) $$ Further factor quadratic: $$x^2 - 2x - 3 = (x - 3)(x + 1)$$ So numerator: $$ (x - 2)(x - 3)(x + 1) $$ 5. **Factor denominator:** $$x^3 - 6x^2 + 11x - 6$$ Try $$x=2$$ as root: $$2^3 - 6(2)^2 + 11(2) - 6 = 0$$ confirms $$x-2$$ is a factor. Divide denominator by $$x-2$$: Coefficients: 1, -6, 11, -6 Bring down 1, multiply by 2: 2, add to -6: -4 Multiply by 2: -8, add to 11: 3 Multiply by 2: 6, add to -6: 0 Quotient: $$x^2 - 4x + 3$$ Factor quadratic: $$x^2 - 4x + 3 = (x - 3)(x - 1)$$ So denominator: $$ (x - 2)(x - 3)(x - 1) $$ 6. **Simplify:** $$f(x) = \frac{(x - 2)(x - 3)(x + 1)}{(x - 2)(x - 3)(x - 1)}$$ Cancel common factors $$x - 2$$ and $$x - 3$$: $$f(x) = \frac{x + 1}{x - 1}$$ 7. **Evaluate limit:** $$\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{x + 1}{x - 1} = \frac{2 + 1}{2 - 1} = \frac{3}{1} = 3$$ **Final answer:** $$\boxed{3}$$