Question

If $f ''\left(0\right)=k, k\ne0,$ then the value of $\displaystyle \lim_{x \to 0}$
$\frac{2f \left(x\right)-3f \left(2x\right)+f \left(4x\right)}{x^{2}}$ is

• A. $k$
• B. $2k$
• C. $3k$
• D. $4k$

Answer 1

Answer: (C)

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