Question

Let $f\left(x\right)$ be a twice-differentiable function and $f^{\prime \prime }\left(0\right)=2$,

then evaluate $\underset{x\to 0}{lim}\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}$.

Answer 1

Given $f^{\prime \prime }\left(0\right)=2$
Consider $\underset{x\to 0}{lim}\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}$
Using L'Hospital's rule
$=\underset{x\to 0}{lim}\frac{2f^{\prime }\left(x\right)-6f^{\prime }\left(2x\right)+4f^{\prime }\left(4x\right)}{2x}$ $\left[as\frac{d}{dx}f\right(x)=f^{\prime }(x),\frac{d}{dx}f(2x)=f^{\prime }(2x\left).2\right]$
Again, applying L'Hospital's rule, we get
$=\underset{x\to 0}{lim}\frac{2f^{\prime \prime }\left(x\right)-12f^{\prime \prime }\left(2x\right)+16f^{\prime \prime }\left(4x\right)}{2}$
On putting $x=0$, we get
$=f^{\prime \prime }\left(0\right)-6f^{\prime \prime }\left(0\right)+8f^{\prime \prime }\left(0\right)$
$=3f^{\prime \prime }\left(0\right)=6$