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If $f^{\prime \prime}(0)=k, k \neq 0,$ then the value of $\lim _{x \rightarrow 0} \frac{2 f(x)-3 f(2 x)+f(4 x)}{x^{2}}$ is
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$3 k$
$\lim _{x \rightarrow 0} \frac{2 f(x)-3 f(2 x)+f(4 x)}{x^{2}}$
$=\lim _{x \rightarrow 0} \frac{2 f^{\prime}(x)-3 f^{\prime}(2 x) \cdot 2+f^{\prime}(4 x) \cdot 4}{2 x}$
$=\lim _{x \rightarrow 0} \frac{f^{\prime}(x)-3 f^{\prime}(2 x)+2 f^{\prime}(4 x)}{x}$
$=\lim _{x \rightarrow 0} \frac{f^{\prime \prime}(x)-3 f''{(2 x)} \cdot 2+2 f^{\prime \prime}(4 x) \cdot 4}{1}$
$=\lim _{x \rightarrow 0} f^{\prime \prime}(x)-6 f''{(2 x)}+8 f''(4 x)$
$=f^{\prime \prime}(0)-6 f^{\prime \prime}(0)+8 f''{(0)}$
$=k-6 k+8 k$
$=3 k$
$=\lim _{x \rightarrow 0} \frac{2 f^{\prime}(x)-3 f^{\prime}(2 x) \cdot 2+f^{\prime}(4 x) \cdot 4}{2 x}$
$=\lim _{x \rightarrow 0} \frac{f^{\prime}(x)-3 f^{\prime}(2 x)+2 f^{\prime}(4 x)}{x}$
$=\lim _{x \rightarrow 0} \frac{f^{\prime \prime}(x)-3 f''{(2 x)} \cdot 2+2 f^{\prime \prime}(4 x) \cdot 4}{1}$
$=\lim _{x \rightarrow 0} f^{\prime \prime}(x)-6 f''{(2 x)}+8 f''(4 x)$
$=f^{\prime \prime}(0)-6 f^{\prime \prime}(0)+8 f''{(0)}$
$=k-6 k+8 k$
$=3 k$
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