Let $f\text{'}\left(x\right)$  be continuous at $x=0$  and $f\text{'}\left(0\right)=k$ , the value of $\underset{x\to 0}{\lim }\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}$  is

$\underset{x\to 0}{\lim }\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2},(\frac{0}{0} form by L\text{'}Hospital rule)$

 $=\underset{x\to 0}{\lim }\frac{2f\text{'}\left(x\right)-6f\text{'}\left(2x\right)+4f\text{'}\left(4x\right)}{2x}, (\frac{0}{0} form by L\text{'}hospital rule)$

   $=\underset{x\to 0}{\lim }\frac{2f\text{'}\text{'}\left(x\right)-12f\text{'}\text{'}\left(2x\right)+16f\text{'}\text{'}\left(4x\right)}{2x},$

$=f\text{'}\text{'}\left(0\right)-6f\text{'}\text{'}\left(0\right)+8f\text{'}\text{'}\left(0\right)=3f\text{'}\text{'}\left(0\right)=3k$ .