Let f(x) = 5 - |x - 2| and g(x) = |x + 1|,whe | vedclass

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(C) $f(x) = 5 - |x - 2|$.$f(x)$ attains its maximum value when $|x - 2| = 0$,which gives $x = 2 = \alpha$.$g(x) = |x + 1|$.$g(x)$ attains its minimum

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$\frac{1}{2}$

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(C) $f(x) = 5 - |x - 2|$.$f(x)$ attains its maximum value when $|x - 2| = 0$,which gives $x = 2 = \alpha$.$g(x) = |x + 1|$.$g(x)$ attains its minimum value when $x + 1 = 0$,which gives $x = -1 = \beta$.We need to evaluate $\lim_{x 	o \alpha \beta} \frac{(x - 1)(x^2 - 5x + 6)}{x^2 - 6x + 8}$.Since $\alpha \beta = (2)(-1) = -2$,the limit is $\lim_{x 	o -2} \frac{(x - 1)(x - 2)(x - 3)}{(x - 2)(x - 4)}$.Canceling the common factor $(x - 2)$,we get $\lim_{x 	o -2} \frac{(x - 1)(x - 3)}{x - 4}$.Substituting $x = -2$,we get $\frac{(-2 - 1)(-2 - 3)}{-2 - 4} = \frac{(-3)(-5)}{-6} = \frac{15}{-6} = -\frac{5}{2}$.Wait,re-evaluating the limit target: $\alpha \beta = -2$. The expression simplifies to $\frac{(x-1)(x-3)}{x-4}$. At $x=-2$,this is $\frac{(-3)(-5)}{-6} = -2.5$. Checking the provided options,there might be a typo in the question's target limit. If the target was $\alpha + \beta = 1$,the limit is $\frac{(0)(-2)}{-3} = 0$. If the target was $\alpha = 2$,the limit is $\frac{(1)(-1)}{-2} = \frac{1}{2}$. Given the options,the intended limit was likely $x 	o \alpha = 2$.

Let $f(x) = 5 - |x - 2|$ and $g(x) = |x + 1|$,where $x \in R$. If $f(x)$ attains its maximum value at $\alpha$ and $g(x)$ attains its minimum value at $\beta$,then $\lim_{x \to \alpha \beta} \frac{(x - 1)(x^2 - 5x + 6)}{x^2 - 6x + 8}$ is equal to:

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