Answer
$-4$
Work Step by Step
Theorem 1
$\displaystyle \lim_{x\rightarrow a}f(x)=L$ if and only if $\displaystyle \lim_{x\rightarrow a^{-}}f(x)=L=\lim_{x\rightarrow a^{+}}f(x)$
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$|2x^{3}-x^{2}|=|x^{2}(2x-1)|=|x^{2}|\cdot|2x-1|$
... the square of a real number is nonnegative, we drop the first absolute bracket,
$|2x^{3}-x^{2}|=x^{2}|2x-1|$
At $x=0.5,\ 2x-1=0.$
$|2x-1|=\left\{\begin{array}{lll}
2x-1 & if & x \geq 0.5\\
-(2x-1) & if & x < 0.5
\end{array}\right.$
So, approaching from the left (for $x < 0.5$)
$\displaystyle \frac{2x-1}{|2x^{3}-x^{2}|}$ =$\displaystyle \frac{2x-1}{x^{2}[-(2x-1)]}=-\frac{(2x-1)}{x^{2}(2x-1)}=-\frac{1}{x^{2}}$
and
$\displaystyle \lim_{x\rightarrow 0.5^{-}}\frac{2x-1}{|2x^{3}-x^{2}|}=\lim_{x\rightarrow 0.5^{-}}-\frac{1}{x^{2}}$
$=-\displaystyle \frac{1}{(0.5)^{2}}=-\frac{1}{0.25}=-4$
