If f(x) is a differentiable function and f''( | vedclass

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(A) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{2f(x) - 3f(2x) + f(4x)}{x^2}$.Since the limit is of the form $\frac{0}{0}$ as $x 	o 0$,we apply $

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$3$

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(A) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{2f(x) - 3f(2x) + f(4x)}{x^2}$.Since the limit is of the form $\frac{0}{0}$ as $x 	o 0$,we apply $L'	ext{Hospital's rule}$.Applying the rule once:$L = \mathop {\lim }\limits_{x 	o 0} \frac{2f'(x) - 3 	imes 2f'(2x) + 4f'(4x)}{2x} = \mathop {\lim }\limits_{x 	o 0} \frac{2f'(x) - 6f'(2x) + 4f'(4x)}{2x}$.Applying $L'	ext{Hospital's rule}$ again:$L = \mathop {\lim }\limits_{x 	o 0} \frac{2f''(x) - 6 	imes 2f''(2x) + 4 	imes 4f''(4x)}{2} = \mathop {\lim }\limits_{x 	o 0} \frac{2f''(x) - 12f''(2x) + 16f''(4x)}{2}$.As $x 	o 0$,$f''(x) 	o f''(0) = a$.$L = \frac{2a - 12a + 16a}{2} = \frac{6a}{2} = 3a$.

If $f(x)$ is a differentiable function and $f''(0) = a$,then $\mathop {\lim }\limits_{x \to 0} \frac{2f(x) - 3f(2x) + f(4x)}{x^2}$ is (in $a$)

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