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(B) We are given the inequality: $\frac{\cos(2x - 4) - 33}{2} < f(x) < \frac{x^2 |4x - 8|}{x - 2}$.First,evaluate the limit of the left-hand side as $

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$-16$

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(B) We are given the inequality: $\frac{\cos(2x - 4) - 33}{2} First,evaluate the limit of the left-hand side as $x 	o 2^-$:$\lim_{x 	o 2^-} \frac{\cos(2x - 4) - 33}{2} = \frac{\cos(0) - 33}{2} = \frac{1 - 33}{2} = \frac{-32}{2} = -16$.Next,evaluate the limit of the right-hand side as $x 	o 2^-$:Since $x $\lim_{x 	o 2^-} \frac{x^2 |4x - 8|}{x - 2} = \lim_{x 	o 2^-} \frac{x^2 (-4(x - 2))}{x - 2} = \lim_{x 	o 2^-} (-4x^2) = -4(2^2) = -16$.By the Sandwich Theorem,since both the left and right limits are $-16$,we have $\lim_{x 	o 2^-} f(x) = -16$.

Let $f : (1, 2) \to R$ satisfy the inequality $\frac{\cos(2x - 4) - 33}{2} < f(x) < \frac{x^2 |4x - 8|}{x - 2}$ for all $x \in (1, 2)$. Then $\lim_{x \to 2^-} f(x)$ is equal to

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