If $f$ is twice differentiable and $f''(0)=2$ then $\lim\limits_{x\to 0}\large\frac{2f(x)-3f(2x)+f(4x)}{x^2}$=

Question

$(a)\;6\qquad(b)\;3\qquad(c)\;12\qquad(d)\;None\;of\;these$

Answer 1

limit=$\lim\limits_{x\to 0}\large\frac{2f'(x)-6f'(2x)+4f'(4x)}{2x}$

$\Rightarrow \lim\limits_{x\to 0}\large\frac{2f''(x)-12f''(2x)+16f''(4x)}{2}$

$\Rightarrow \large\frac{6f''(0)}{2}$

$\Rightarrow 6$

Hence (a) is the correct answer.