If f^{prime prime}(0)=k, k neq 0, then the va | vedclass

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(C) Given the limit $L = \lim _{x \rightarrow 0} \frac{2 f(x)-3 f(2 x)+f(4 x)}{x^{2}}$.Since the limit is of the form $\frac{0}{0}$, we apply $L$'$H$ô

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$3 k$

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(C) Given the limit $L = \lim _{x \rightarrow 0} \frac{2 f(x)-3 f(2 x)+f(4 x)}{x^{2}}$.Since the limit is of the form $\frac{0}{0}$, we apply $L$'$H$ôpital's rule:$L = \lim _{x \rightarrow 0} \frac{2 f^{\prime}(x)-6 f^{\prime}(2 x)+4 f^{\prime}(4 x)}{2 x}$$L = \lim _{x \rightarrow 0} \frac{f^{\prime}(x)-3 f^{\prime}(2 x)+2 f^{\prime}(4 x)}{x}$Applying $L$'$H$ôpital's rule again:$L = \lim _{x}$ ${\rightarrow 0} \frac{f^{\prime \prime}(x)-3 f^{\prime \prime}(2 x) \cdot 2+2 f^{\prime \prime}(4 x) \cdot 4}{1}$$L = f^{\prime \prime}(0)-6 f^{\prime \prime}(0)+8 f^{\prime \prime}(0)$$L = k-6 k+8 k = 3 k$.

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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