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

given f ( x ) is continuous at x=0 =limx→0 f′′(x)=f′′(0)=4 Now, limx→0 2f(x)−3f(2x)+f(4x)x200 form =limx→0 2f′(x)−6f′(2x)+4f′(4x)2x00 form ∣=limx→0 2f′′(x)−12f′′(x)+16f′′(4x)2[Using L' Hospital Rule successively] =2f′′(0)−12f′′(0)+16f′′(0)2=12

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

given f ( x ) is continuous at x=0 =limx→0 f′′(x)=f′′(0)=4 Now, limx→0 2f(x)−3f(2x)+f(4x)x200 form =limx→0 2f′(x)−6f′(2x)+4f′(4x)2x00 form ∣=limx→0 2f′′(x)−12f′′(x)+16f′′(4x)2[Using L' Hospital Rule successively] =2f′′(0)−12f′′(0)+16f′′(0)2=12

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

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Detailed Solution

given f ( x ) is continuous at $x = 0$

$= lim_{x \to 0} f^{''} (x) = f^{''} (0) = 4$

Now, $lim_{x \to 0} \frac{2 f (x) - 3 f (2 x) + f (4 x)}{x^{2}} [\frac{0}{0} form]$

$\begin{array}{l} = lim_{x \to 0} \frac{2 f^{'} (x) - 6 f^{'} (2 x) + 4 f^{'} (4 x)}{2 x} [\frac{0}{0} form ∣] \\ = lim_{x \to 0} \frac{2 f^{''} (x) - 12 f^{''} (x) + 16 f^{''} (4 x)}{2} \end{array}$