Question

Let $f:(1,2)\to R$ satisfy the inequality
$\frac{\cos (2x-4)-33}{2}<f\left(x\right)<\frac{x^2|4x-8|}{x-2},\forall x\in (1,2).$ Then $\underset{x\to 2^{-}}{lim}f\left(x\right)$ is

• A. $16$
• B. $-16$
• C. $8$
• D. does not exist

Answer 1

The correct option is B $-16$

$\underset{x\to 2^{-}}{lim}\frac{\cos (2x-4)-33}{2}=-16$ and
$\underset{x\to 2^{-}}{lim}\frac{x^2|4x-8|}{x-2}=-4\left(4\right)=-16$
By sandwich theorem, $\underset{x\to 2^{-}}{lim}f\left(x\right)=-16$