Define f(x) = begin{cases} frac{1-sin x}{(pi- | vedclass

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(B) For $f(x)$ to be continuous at $x = \frac{\pi}{2}$,we must have $\lim_{x 	o \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}ight) = k$. Let $x = \fra

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$\frac{1}{8}$

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(B) For $f(x)$ to be continuous at $x = \frac{\pi}{2}$,we must have $\lim_{x 	o \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}ight) = k$. Let $x = \frac{\pi}{2} + h$. As $x 	o \frac{\pi}{2}$,$h 	o 0$. Then $\pi - 2x = \pi - 2(\frac{\pi}{2} + h) = -2h$. Also,$\sin x = \sin(\frac{\pi}{2} + h) = \cos h$. Substituting these into the limit: $\lim_{h 	o 0} \frac{1 - \cos h}{(-2h)^2} = \lim_{h 	o 0} \frac{1 - \cos h}{4h^2}$. Using the identity $1 - \cos h = 2 \sin^2(\frac{h}{2})$,we get: $\lim_{h 	o 0} \frac{2 \sin^2(\frac{h}{2})}{4h^2} = \lim_{h 	o 0} \frac{1}{2} \left( \frac{\sin(h/2)}{h} ight)^2 = \lim_{h 	o 0} \frac{1}{2} \left( \frac{\sin(h/2)}{2(h/2)} ight)^2 = \frac{1}{2} 	imes \left( \frac{1}{2} ight)^2 = \frac{1}{2} 	imes \frac{1}{4} = \frac{1}{8}$. Thus,$k = \frac{1}{8}$.

Define $f(x) = \begin{cases} \frac{1-\sin x}{(\pi-2x)^2} & \text{, if } x \neq \frac{\pi}{2} \\ k & \text{, if } x = \frac{\pi}{2} \end{cases}$. If $f(x)$ is continuous at $x = \frac{\pi}{2}$,then $k =$

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