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$Let f^{'} (x) be continuous at x = 0 and f^{''} (0) = 4 . The value of$ $\lim_{x \to 0} \frac{2 f (x) - 3 f (2 x) + f (4 x)}{x^{2}}$

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Correct option is C

limx→02f(x)−3f(2x)+f(4x)x2Form 00=limx→02f'(x)−6f'(2x)+4f'(4x)2xForm 00=limx→02f''x−12f"2x+16f"4x=f"0−6f"0+gf"0=3f"0=3×4=12Hence (3)is the correct answer.

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Let f′(x) be continuous at x=0 and f′′(0)=4. The value of limx→02f(x)−3f(2x)+f(4x)x2

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$Let f^{'} (x) be continuous at x = 0 and f^{''} (0) = 4 . The value of$ $\lim_{x \to 0} \frac{2 f (x) - 3 f (2 x) + f (4 x)}{x^{2}}$