If the function f defined on left( frac{pi }{ | vedclass

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(C) For the function to be continuous at $x = \frac{\pi}{4}$,the limit of $f(x)$ as $x 	o \frac{\pi}{4}$ must equal $f\left(\frac{\pi}{4}ight)$.$\l

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$\frac{1}{2}$

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(C) For the function to be continuous at $x = \frac{\pi}{4}$,the limit of $f(x)$ as $x 	o \frac{\pi}{4}$ must equal $f\left(\frac{\pi}{4}ight)$.$\lim_{x 	o \frac{\pi}{4}} f(x) = f\left(\frac{\pi}{4}ight) = k$$\lim_{x 	o \frac{\pi}{4}} \frac{\sqrt{2} \cos x - 1}{\cot x - 1} = k$Since this is a $\frac{0}{0}$ form,we apply $L$'Hospital's Rule:$\lim_{x 	o \frac{\pi}{4}} \frac{\frac{d}{dx}(\sqrt{2} \cos x - 1)}{\frac{d}{dx}(\cot x - 1)} = k$$\lim_{x 	o \frac{\pi}{4}} \frac{-\sqrt{2} \sin x}{-\csc^2 x} = k$$\lim_{x 	o \frac{\pi}{4}} \sqrt{2} \sin x \cdot \sin^2 x = k$$\lim_{x 	o \frac{\pi}{4}} \sqrt{2} \sin^3 x = k$Substituting $x = \frac{\pi}{4}$:$k = \sqrt{2} \left(\frac{1}{\sqrt{2}}ight)^3 = \sqrt{2} \cdot \frac{1}{2\sqrt{2}} = \frac{1}{2}$

If the function $f$ defined on $\left( \frac{\pi }{6}, \frac{\pi }{3} \right)$ by $f(x) = \begin{cases} \frac{\sqrt{2} \cos x - 1}{\cot x - 1}, & x \neq \frac{\pi}{4} \\ k, & x = \frac{\pi}{4} \end{cases}$ is continuous,then $k$ is equal to

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