Question

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

Answer 1

Answer: (C)

$\underset{x\to 0}{\lim }\frac{2f(x)-3f(2x)+f(4x)}{x^2}$
$=\underset{x\to 0}{\lim }\frac{2f^{\prime }(x)-3f^{\prime }(2x)\cdot 2+f^{\prime }(4x)\cdot 4}{2x}$
$=\underset{x\to 0}{\lim }\frac{f^{\prime }(x)-3f^{\prime }(2x)+2f^{\prime }(4x)}{x}$
$=\underset{x\to 0}{\lim }\frac{f^{\prime \prime }(x)-3f^{\prime \prime }(2x)\cdot 2+2f^{\prime \prime }(4x)\cdot 4}{1}$
$=\underset{x\to 0}{\lim }{f}^{\prime \prime }(x)-6f^{\prime \prime }(2x)+8f^{\prime \prime }(4x)$
$=f^{\prime \prime }(0)-6f^{\prime \prime }(0)+8f^{\prime \prime }(0)$
$=k-6k+8k\left[\because f^{\prime \prime }(0)=k\right]$
$=3k$