Question

If $f^{\prime \prime }\left(0\right)=k$, 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. $k$
• B. $2k$
• C. $3k$
• D. $4k$

Answer 1

The correct option is C $3k$

$f^{\prime \prime }\left(0\right)=k$, (given)

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

Put $x=0$ we get $\frac{0}{0}$ form

So, apply L'Hospital's Rule

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

again put $x=0$, we get $\frac{0}{0}$, apply L'Hospital's again

$\therefore$ $\underset{x\to 0}{lim}$ $\frac{2f^{\prime \prime }\left(x\right)-12f^{\prime \prime }\left(2x\right)+16f^{\prime \prime }\left(4x\right)}{2}$

$=\frac{2f^{\prime \prime }\left(0\right)-12f^{\prime \prime }\left(0\right)+16f^{\prime \prime }\left(0\right)}{2}$

$=\frac{6f^{\prime \prime }\left(0\right)}{2}$

$=3f^{\prime \prime }\left(0\right)$

$=3k$