If f''(x) is continuous at x = 0 and f''(0) = | vedclass

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(C) Given the limit $L = \lim_{x 	o 0} \frac{2f(x) - 3f(2x) + f(4x)}{x^2}$.Substituting $x = 0$ gives the $\frac{0}{0}$ indeterminate form.Applying $

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$12$

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(C) Given the limit $L = \lim_{x 	o 0} \frac{2f(x) - 3f(2x) + f(4x)}{x^2}$.Substituting $x = 0$ gives the $\frac{0}{0}$ indeterminate form.Applying $L$'Hospital's rule,we differentiate the numerator and denominator with respect to $x$:$L = \lim_{x 	o 0} \frac{2f'(x) - 3 \cdot 2f'(2x) + 4f'(4x)}{2x} = \lim_{x 	o 0} \frac{2f'(x) - 6f'(2x) + 4f'(4x)}{2x}$.Applying $L$'Hospital's rule again:$L = \lim_{x 	o 0} \frac{2f''(x) - 6 \cdot 2f''(2x) + 4 \cdot 4f''(4x)}{2} = \lim_{x 	o 0} \frac{2f''(x) - 12f''(2x) + 16f''(4x)}{2}$.Since $f''(x)$ is continuous at $x = 0$,we have $\lim_{x 	o 0} f''(x) = f''(0) = 4$.Substituting $x = 0$ into the expression:$L = \frac{2f''(0) - 12f''(0) + 16f''(0)}{2} = \frac{6f''(0)}{2} = 3f''(0)$.Given $f''(0) = 4$,we get $L = 3 	imes 4 = 12$.

If $f''(x)$ is continuous at $x = 0$ and $f''(0) = 4$,then find the value of $\lim_{x \to 0} \frac{2f(x) - 3f(2x) + f(4x)}{x^2}$.

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