1. **Problem:** Show that $$\lim_{x \to 2} \left[\frac{x^3 - 2x^2 + 2x - 4}{x^2 - 5x + 6}\right] = -6$$ 2. **Formula and rules:** To find limits of rational functions, if direct substitution leads to an indeterminate form like $$\frac{0}{0}$$, factor numerator and denominator and simplify. 3. **Step 1: Substitute $$x=2$$ directly:** $$\frac{2^3 - 2 \cdot 2^2 + 2 \cdot 2 - 4}{2^2 - 5 \cdot 2 + 6} = \frac{8 - 8 + 4 - 4}{4 - 10 + 6} = \frac{0}{0}$$ which is indeterminate. 4. **Step 2: Factor numerator and denominator:** - Numerator: $$x^3 - 2x^2 + 2x - 4$$ Group terms: $$ (x^3 - 2x^2) + (2x - 4) = x^2(x - 2) + 2(x - 2) = (x - 2)(x^2 + 2)$$ - Denominator: $$x^2 - 5x + 6 = (x - 2)(x - 3)$$ 5. **Step 3: Simplify the expression:** $$\frac{(x - 2)(x^2 + 2)}{(x - 2)(x - 3)} = \frac{x^2 + 2}{x - 3}, \quad x \neq 2$$ 6. **Step 4: Take the limit as $$x \to 2$$:** $$\lim_{x \to 2} \frac{x^2 + 2}{x - 3} = \frac{2^2 + 2}{2 - 3} = \frac{4 + 2}{-1} = \frac{6}{-1} = -6$$ 7. **Answer:** $$\boxed{-6}$$ This completes the proof that the limit is $$-6$$.