Question

Let $f:(1,2)\to R$ satisfies the equality
$\frac{\cos (2x-4)-33}{2}<f\left(x\right)<\frac{x^2|4x-8|}{x-2}$,$\forall$ $x$ $ϵ(1,2)$.Then ${lim}_{x\to 2}f\left(x\right)$ is equal to

• A. $16$
• B. $-16$
• C. cannot be determined from the given information
• D. does not exists

Answer 1

Answer: (B)

$-16$

Since, for $x\in (1,2)$

$\frac{\cos (2x-4)-33}{2}<f\left(x\right)<\frac{x^2|4x-8|}{x-2}$

$\underset{x\to 2}{lim}\frac{\cos (2x-4)-33}{2}<\underset{x\to 2}{lim}f\left(x\right)<\underset{x\to 2}{lim}\frac{x^2|4x-8|}{x-2}$

$\underset{x\to 2}{lim}\frac{\cos (2x-4)-33}{2}=\frac{1-33}{2}=-16$

$\underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{-}}{lim}\frac{x^2|4x-8|}{x-2}=\underset{h\to 0}{lim}\frac{(2-h{)}^2|-4h|}{-h}=-16$

So, by Sandwich theorem,

$\underset{x\to 2}{lim}f\left(x\right)=-16$

Hence, option 'B' is correct.