If f(x) = frac{4}{x^4} left[ 1 - cos frac{x}{ | vedclass

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(D) Given $f(x) = \frac{4}{x^4} \left[ 1 - \cos \frac{x}{2} - \cos \frac{x}{4} + \cos \frac{x}{2} \cdot \cos \frac{x}{4} \right]$.Factoring the expres

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$\frac{1}{64}$

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(D) Given $f(x) = \frac{4}{x^4} \left[ 1 - \cos \frac{x}{2} - \cos \frac{x}{4} + \cos \frac{x}{2} \cdot \cos \frac{x}{4} ight]$.Factoring the expression inside the brackets:$f(x) = \frac{4}{x^4} \left[ (1 - \cos \frac{x}{2}) - \cos \frac{x}{4} (1 - \cos \frac{x}{2}) ight] = \frac{4}{x^4} (1 - \cos \frac{x}{2}) (1 - \cos \frac{x}{4})$.Since $f(x)$ is continuous at $x = 0$,$f(0) = \lim_{x 	o 0} f(x)$.Using the identity $1 - \cos 	heta = 2 \sin^2 \frac{	heta}{2}$:$f(0) = \lim_{x 	o 0} \frac{4}{x^4} \left( 2 \sin^2 \frac{x}{4} ight) \left( 2 \sin^2 \frac{x}{8} ight) = 16 \lim_{x 	o 0} \frac{\sin^2 \frac{x}{4}}{x^2} \cdot \frac{\sin^2 \frac{x}{8}}{x^2}$.Multiply and divide by $(\frac{1}{4})^2$ and $(\frac{1}{8})^2$:$f(0) = 16 \lim_{x 	o 0} \left( \frac{\sin \frac{x}{4}}{\frac{x}{4}} ight)^2 \cdot \frac{1}{16} \cdot \left( \frac{\sin \frac{x}{8}}{\frac{x}{8}} ight)^2 \cdot \frac{1}{64} = 16 \cdot \frac{1}{16} \cdot \frac{1}{64} \cdot (1)^2 \cdot (1)^2 = \frac{1}{64}$.

If $f(x) = \frac{4}{x^4} \left[ 1 - \cos \frac{x}{2} - \cos \frac{x}{4} + \cos \frac{x}{2} \cdot \cos \frac{x}{4} \right]$ is continuous at $x = 0$,then $f(0)$ is

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