Question

Let $f\left(x\right)$ be twice differentiable function such that $f^{\prime \prime }\left(0\right)=2$. Then, $\underset{x\to 0}{lim}\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}$ is equal to.

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

Answer 1

Answer: (A)

$6$

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

Using L' Hospital's rule, we get
$=\underset{x\to 0}{lim}\frac{2f^{\prime }\left(x\right)-6f^{\prime }\left(2x\right)+4f^{\prime }\left(4x\right)}{2x}$ ....... $\left[\frac{0}{0}form\right]$

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