Let f(x) be a twice differentiable funciton and f¹¹(0)=2, then $\underset{x\to 0}{Lim}\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}$ is

$\begin{array}{l}f"\left(0\right)=2\\ \underset{x\to 0}{Lt}\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}\left(\frac{0}{0} formula\right)\\ \underset{x\to a}{Lt}\frac{2f\text{'}\left(x\right)-3f\text{'}\left(2x\right)2+f\text{'}\left(4x\right).4}{2x}\left(\frac{0}{0}formula\right)\\ \underset{x\to 0}{Lt}\frac{2f"\left(x\right)-6f\left(2x\right)2+4f"\left(4x\right)4}{2}\left(\text{L'\hspace{0.17em}}hosiptalrule\right)\\ \frac{2f"\left(0\right)-12f"\left(0\right)+16f"\left(0\right)}{2}=\frac{2\left(2-12+16\right)}{2}=6\end{array}$