Let $f\left(x\right)$ be twice differentiable function such that $f^{\prime \prime }\left(0\right)=2$. Then, $\underset{x\to 0}{lim} \frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}$, is

Solution:

We have,

$\underset{x\to 0}{lim} \frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right) }{x^2} \left[\frac{0}{0}\text{ form }\right]$

$=\underset{x\to 0}{lim} \frac{2f^{\mathrm{\prime }}\left(x\right)-6f^{\mathrm{\prime }}\left(2x\right)+4f^{\mathrm{\prime }}\left(4x\right)}{2x}$     

                                                   [Using L' Hospital's Rule] 

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

                                                   [Using L' Hospital's Rule] 

$\begin{array}{l}= \frac{2f^{\prime \prime }\left(0\right)-12f^{\prime \prime }\left(0\right)+16f^{\prime \prime }\left(0\right)}{2}\\ = 3f^{\prime \prime }\left(0\right)=3\times 2=6\end{array}$