Question

If $f^n\left(x\right)$ be continuous at $x=0$ and $f^n\left(0\right)=4$ then $\underset{x\to 0}{lim}\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}=$

Answer 1

Let $L=\underset{x\to 0}{lim}\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}$
Clearly form of the limit is, $\frac{0}{0}$
Thus applying L-Hospital's rule,
$L=\underset{x\to 0}{lim}\frac{2f^{\prime }\left(x\right)-6f^{\prime }\left(2x\right)+4f^{\prime }\left(4x\right)}{2x}$ still $\frac{0}{0}$ form
So again applying L -Hospital's rule,
$L=\underset{x\to 0}{lim}\frac{2f^{\prime \prime }\left(x\right)-12f^{\prime \prime }\left(2x\right)+16f^{\prime \prime }\left(4x\right)}{2}=3f^{\prime \prime }\left(0\right)=12$