If the function f(x) = frac{1 - cos 4x}{8x^2} | vedclass

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(B) For a function $f(x)$ to be continuous at $x = 0$,the limit of $f(x)$ as $x 	o 0$ must equal the value of the function at $x = 0$,i.e.,$\lim_{x \

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$1$

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(B) For a function $f(x)$ to be continuous at $x = 0$,the limit of $f(x)$ as $x 	o 0$ must equal the value of the function at $x = 0$,i.e.,$\lim_{x 	o 0} f(x) = f(0) = k$.Given $f(x) = \frac{1 - \cos 4x}{8x^2}$ for $x 
e 0$,we calculate the limit:$\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \frac{1 - \cos 4x}{8x^2}$Using the trigonometric identity $1 - \cos 	heta = 2 \sin^2(\frac{	heta}{2})$,we have $1 - \cos 4x = 2 \sin^2(2x)$.Substituting this into the limit:$\lim_{x 	o 0} \frac{2 \sin^2(2x)}{8x^2} = \lim_{x 	o 0} \frac{\sin^2(2x)}{4x^2}$This can be rewritten as:$\lim_{x 	o 0} \left( \frac{\sin 2x}{2x} ight)^2$Since $\lim_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$,we have:$(1)^2 = 1$Thus,$k = 1$.

If the function $f(x) = \frac{1 - \cos 4x}{8x^2}$ for $x \ne 0$ and $f(x) = k$ for $x = 0$ is a continuous function at $x = 0$,then the value of $k$ is:

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