Key Concepts and Formulas

• Absolute Value Function Properties: The expression $|a|$ represents the distance of $a$ from zero. Thus, $|a| \ge 0$ for all real $a$. The minimum value of $|a|$ is 0, which occurs when $a=0$.
• Maximizing/Minimizing Functions with Absolute Values: To find the maximum or minimum of a function involving absolute values, analyze the behavior of the absolute value term. For example, to maximize $C - |x-a|$, we need to minimize $|x-a|$. To minimize $C + |x-a|$, we need to minimize $|x-a|$.
• Limit Evaluation Techniques: For rational functions where direct substitution leads to an indeterminate form (like $\frac{0}{0}$), factorize the numerator and denominator and cancel out common factors.

Step-by-Step Solution

Step 1: Determine the value of $\alpha$. We are given the function $f(x) = 5 - |x - 2|$. To find the value of $x$ where $f(x)$ attains its maximum value, we need to minimize the term $|x - 2|$. The absolute value $|x - 2|$ is always non-negative, and its minimum value is 0. This minimum occurs when $x - 2 = 0$, which means $x = 2$. Therefore, $f(x)$ attains its maximum value when $x = 2$. So, $\alpha = 2$.

Step 2: Determine the value of $\beta$. We are given the function $g(x) = |x + 1|$. To find the value of $x$ where $g(x)$ attains its minimum value, we need to find the value of $x$ that minimizes $|x + 1|$. The absolute value $|x + 1|$ is always non-negative, and its minimum value is 0. This minimum occurs when $x + 1 = 0$, which means $x = -1$. Therefore, $g(x)$ attains its minimum value when $x = -1$. So, $\beta = -1$.

Step 3: Calculate the value of $\alpha \beta$. Using the values found in Step 1 and Step 2, we calculate the product $\alpha \beta$. $\alpha \beta = (2)(-1) = -2$.

Step 4: Evaluate the limit. We need to evaluate the limit: \mathop {\lim }\limits_{x \to -\alpha \beta } {{\left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right)} \over {{x^2} - 6x + 8}}} Substitute the value of $\alpha \beta = -2$ into the limit expression: $\mathop {\lim }\limits_{x \to -(-2)} {{\left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right)} \over {{x^2} - 6x + 8}} = \mathop {\lim }\limits_{x \to 2} {{\left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right)} \over {{x^2} - 6x + 8}}$ Now, we need to factorize the quadratic expressions in the numerator and the denominator. For the numerator, $x^2 - 5x + 6$: We look for two numbers that multiply to 6 and add up to -5. These numbers are -2 and -3. So, $x^2 - 5x + 6 = (x - 2)(x - 3)$. For the denominator, $x^2 - 6x + 8$: We look for two numbers that multiply to 8 and add up to -6. These numbers are -2 and -4. So, $x^2 - 6x + 8 = (x - 2)(x - 4)$.

Substitute these factorizations back into the limit expression: $\mathop {\lim }\limits_{x \to 2} {{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)} \over {\left( {x - 2} \right)\left( {x - 4} \right)}}$ As $x \to 2$, $x \neq 2$, so we can cancel out the common factor $(x - 2)$ from the numerator and the denominator: $\mathop {\lim }\limits_{x \to 2} {{\left( {x - 1} \right)\left( {x - 3} \right)} \over {\left( {x - 4} \right)}}$ Now, we can directly substitute $x = 2$ into the simplified expression, as the denominator will not be zero: ${{\left( {2 - 1} \right)\left( {2 - 3} \right)} \over {\left( {2 - 4} \right)}} = {{\left( 1 \right)\left( -1 \right)} \over {\left( -2 \right)}} = {-1 \over -2} = {1 \over 2}$

Common Mistakes & Tips

• Incorrectly identifying $\alpha$ and $\beta$: Ensure you correctly understand how to maximize/minimize functions involving absolute values. For $C - |expression|$, maximum occurs when $|expression|$ is minimum. For $|expression|$, minimum occurs when $expression = 0$.
• Algebraic errors during factorization: Double-check your factorization of quadratic expressions. A simple mistake here can lead to an incorrect final answer.
• Not canceling common factors: When evaluating limits of rational functions that result in $\frac{0}{0}$ form, failing to factorize and cancel common factors will prevent you from finding the correct limit.

Summary The problem requires us to first find the values of $\alpha$ and $\beta$ by analyzing the maximum of $f(x)$ and the minimum of $g(x)$. The function $f(x) = 5 - |x - 2|$ attains its maximum when $|x-2|$ is minimum, which is at $x=2$, so $\alpha=2$. The function $g(x) = |x + 1|$ attains its minimum when $|x+1|$ is minimum, which is at $x=-1$, so $\beta=-1$. Then, we need to evaluate the limit as $x$ approaches $\alpha\beta = -2$. After substituting $\alpha\beta$, the limit becomes \mathop {\lim }\limits_{x \to 2} {{\left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right)} \over {{x^2} - 6x + 8}}}. By factorizing the quadratic terms and canceling the common factor $(x-2)$, we simplify the expression and evaluate the limit by direct substitution, yielding $\frac{1}{2}$.

The final answer is $\boxed{\frac{1}{2}}$.