Question

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

• A. $6$
• B. $3$
• C. $12$
• D. none of these

Answer 1

The correct option is A $6$

By L'Hospital's Rule,
$=\underset{x\to 0}{lim}\frac{2f^{\prime }\left(x\right)-6f^{\prime }\left(2x\right)+4f^{\prime }\left(4x\right)}{2x}$
$=\underset{x\to 0}{lim}\frac{2f^{\prime \prime }\left(x\right)-12f^{\prime \prime }\left(2x\right)+16f^{\prime \prime }\left(4x\right)}{2}$
$=\frac{4-24+32}{2}$
$=\frac{12}{2}=6$