The value of f(0) so that the function f(x) = | vedclass

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(D) For the function $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x 	o 0} f(x)$.Using the identity $1 - \cos 	heta = 2 \sin^2(\frac

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$\frac{1}{8}$

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(D) For the function $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x 	o 0} f(x)$.Using the identity $1 - \cos 	heta = 2 \sin^2(\frac{	heta}{2})$,we have:$\lim_{x 	o 0} \frac{1 - \cos(1 - \cos x)}{x^4} = \lim_{x 	o 0} \frac{2 \sin^2(\frac{1 - \cos x}{2})}{x^4}$Since $1 - \cos x = 2 \sin^2(\frac{x}{2})$,this becomes:$\lim_{x 	o 0} \frac{2 \sin^2(\sin^2(\frac{x}{2}))}{x^4}$Using the limit $\lim_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$,we approximate $\sin 	heta \approx 	heta$ for small $	heta$:$\lim_{x 	o 0} \frac{2 (\sin^2(\frac{x}{2}))^2}{x^4} = \lim_{x 	o 0} \frac{2 (\frac{x}{2})^4}{x^4} = \lim_{x 	o 0} \frac{2 \cdot \frac{x^4}{16}}{x^4} = \frac{2}{16} = \frac{1}{8}$.

The value of $f(0)$ so that the function $f(x) = \frac{1 - \cos(1 - \cos x)}{x^4}$ is continuous everywhere is

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