Question

$f^{\prime }\left(x\right)$ is continuous at $x=0$ and $f^{\prime \prime }\left(0\right)=4$ then the value of $\underset{x\to 0}{lim}\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}$ is

• A. $4$
• B. $8$
• C. $12$
• D. None of these

Answer 1

Answer: (C)

$12$

$\underset{x\to 0}{lim}\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}$
$=\underset{x\to 0}{lim}\frac{2f^{\prime }\left(x\right)-6f^{\prime }\left(2x\right)+4f^{\prime }\left(4x\right)}{2x}$, using L-Hospital's rule
$=\underset{x\to 0}{lim}\frac{2f^{\prime \prime }\left(x\right)-12f^{\prime \prime }\left(2x\right)+16f^{\prime \prime }\left(4x\right)}{2}$,again using L-Hospital's rule
$=\frac{2f^{\prime \prime }\left(0\right)-12f^{\prime \prime }\left(0\right)+16f^{\prime \prime }\left(0\right)}{2}=3f^{\prime \prime }\left(0\right)=12$