Question

Let $f:(1,2)\to R$ satisfies 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 find $\underset{x\to 2^{-}}{lim}f\left(x\right)$

Answer 1

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

$\Rightarrow$$\underset{x\to 2^{-}}{lim}\frac{\cos (2x-4)-33}{2}\le$ $\underset{x\to 2^{-}}{lim}f\left(x\right)$ $\le \underset{x\to 2^{-}}{lim}\frac{x^2|4x-8|}{x-2}$ [ Using limit property]

or, $\underset{x\to 2^{-}}{lim}\frac{\cos (2x-4)-33}{2}\le$ $\underset{x\to 2^{-}}{lim}f\left(x\right)$ $\le \underset{x\to 2^{-}}{lim}(-4x^2)$ [Since, $|4x-8|=4|x-2|=-4(x-2)$ when $x<2$. ]

or, $\frac{1-33}{2}\le$ $\underset{x\to 2^{-}}{lim}f\left(x\right)$ $\le (-{4.2}^2)$

or, $-16\le$ $\underset{x\to 2^{-}}{lim}f\left(x\right)$ $\le -16$

$\Rightarrow$$\underset{x\to 2^{-}}{lim}f\left(x\right)=-16$[ By Sandwich property].