Question

If $f^{\prime }$ is differentiable function and $f^{\prime \prime }\left(x\right)$ is continuous at $x=0$ and $f^{\prime \prime }\left(0\right)=a$, 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. $a$
• B. $2a$
• C. $3a$
• D. none of these

Answer 1

The correct option is C $3a$

We know that, by L-Hospital's rule,
$\underset{x\to 0}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to 0}{lim}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}=\frac{f^{\prime }\left(0\right)}{g^{\prime }\left(0\right)}$, if $f\left(0\right)=g\left(0\right)=0$
$\therefore$ $\underset{x\to 0}{lim}\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}$

$=\underset{x\to 0}{lim}\frac{2f^{\prime }\left(x\right)-6f^{\prime }\left(2x\right)+4f^{\prime }\left(4x\right)}{2x}$
$=\underset{x\to 0}{lim}\frac{f^{\prime }\left(x\right)-3f^{\prime }\left(2x\right)+2f^{\prime }\left(4x\right)}{x}$

$=\underset{x\to 0}{lim}(f^{\prime \prime }\left(x\right)-6f^{\prime \prime }\left(2x\right)+8f^{\prime \prime }\left(4x\right))$
$=f^{\prime \prime }\left(0\right)-6f^{\prime \prime }\left(0\right)+8f^{\prime \prime }\left(0\right)=a-6a+8a=3a$