Question

$x\stackrel{lim}{\to }0\frac{f\left(4x\right)-3f\left(3x\right)+f\left(2x\right)+f\left(x\right)}{x^3}=12$
$f^{\prime }\left(x\right)=0$ and $f^{\prime \prime }\left(x\right)=0$. Then find $f^{\prime \prime \prime }\left(0\right)$

Answer 1

${lim}_{x\to 0}\frac{f\left(4x\right)-3f\left(3x\right)+f\left(2x\right)+f\left(x\right)}{x^3}=12$

Consider the $L.H.S$

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

Putting the limits we get $\frac{0}{0}$ form, Hence using L' Hospitals rule

$={lim}_{x\to 0}\frac{4f^{{}^{\prime }}\left(4x\right)-9f^{{}^{\prime }}\left(3x\right)+2f^{{}^{\prime }}\left(2x\right)+f^{{}^{\prime }}\left(x\right)}{3x^2}={lim}_{x\to 0}\frac{16f^{{}^{\prime \prime }}\left(4x\right)-27f^{{}^{\prime \prime }}\left(3x\right)+4f^{{}^{\prime \prime }}\left(2x\right)+f^{{}^{\prime \prime }}\left(x\right)}{6x}={lim}_{x\to 0}\frac{64f^{{}^{\prime \prime \prime }}\left(4x\right)-81f^{{}^{\prime \prime \prime }}\left(3x\right)+8f^{{}^{\prime \prime \prime }}\left(x\right)+f^{{}^{\prime \prime \prime }}\left(x\right)}{6}$

Putting the limits

$=\frac{-8f^{{}^{\prime \prime \prime }}\left(0\right)}{6}$

Equating LHS with RHS we get

$\frac{-8f^{{}^{\prime \prime \prime }}\left(0\right)}{6}=12$

$\Rightarrow f^{{}^{\prime \prime \prime }}\left(0\right)=-9$