Let f(x) be twice differentiable function such that f′′(0)=2. Then, limx→0 2f(x)−3f(2x)+f(4x)x2, is

We have,

$lim_{x \to 0} \frac{2 f (x) - 3 f (2 x) + f (4 x)}{x^{2}} [\frac{0}{0} form]$

$= lim_{x \to 0} \frac{2 f^{'} (x) - 6 f^{'} (2 x) + 4 f^{'} (4 x)}{2 x}$

[Using L' Hospital's Rule]

$= lim_{x \to 0} \frac{2 f^{''} (x) - 12 f^{''} (2 x) + 16 f^{''} (4 x)}{2}$

[Using L' Hospital's Rule]

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

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