If the function f(x) = begin{cases} frac{sqrt | vedclass

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(D) For the function $f(x)$ to be continuous at $x = \pi$,we must have $\lim_{x 	o \pi} f(x) = f(\pi) = k$.Let $x = \pi + h$,where $h 	o 0$ as $x

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$0.25$

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(D) For the function $f(x)$ to be continuous at $x = \pi$,we must have $\lim_{x 	o \pi} f(x) = f(\pi) = k$.Let $x = \pi + h$,where $h 	o 0$ as $x 	o \pi$. Then $(\pi - x)^2 = (-h)^2 = h^2$.$\lim_{h 	o 0} \frac{\sqrt{2 + \cos(\pi + h)} - 1}{h^2} = k$.Since $\cos(\pi + h) = -\cos h$,we have:$k = \lim_{h 	o 0} \frac{\sqrt{2 - \cos h} - 1}{h^2}$.Rationalizing the numerator:$k = \lim_{h 	o 0} \frac{(\sqrt{2 - \cos h} - 1)(\sqrt{2 - \cos h} + 1)}{h^2(\sqrt{2 - \cos h} + 1)} = \lim_{h 	o 0} \frac{2 - \cos h - 1}{h^2(\sqrt{2 - \cos h} + 1)}$.$k = \lim_{h 	o 0} \frac{1 - \cos h}{h^2(\sqrt{2 - \cos h} + 1)}$.Using the identity $1 - \cos h = 2 \sin^2(h/2)$:$k = \lim_{h 	o 0} \frac{2 \sin^2(h/2)}{h^2(\sqrt{2 - \cos h} + 1)} = \lim_{h 	o 0} \frac{2 \sin^2(h/2)}{4(h/2)^2(\sqrt{2 - \cos h} + 1)}$.Since $\lim_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$:$k = \frac{2}{4} 	imes \frac{1}{\sqrt{2 - 1} + 1} = \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{4} = 0.25$.

If the function $f(x) = \begin{cases} \frac{\sqrt{2 + \cos x} - 1}{(\pi - x)^2}, & x \neq \pi \\ k, & x = \pi \end{cases}$ is continuous at $x = \pi$,then $k$ equals:

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