If f(x) = begin{cases} 1, & 0

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(C) To check the continuity at $x = \frac{3\pi}{4}$,we evaluate the left-hand limit,right-hand limit,and the value of the function at that point.First

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$f(x)$ is continuous at $x = \frac{3\pi}{4}$

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(C) To check the continuity at $x = \frac{3\pi}{4}$,we evaluate the left-hand limit,right-hand limit,and the value of the function at that point.First,the value of the function at $x = \frac{3\pi}{4}$ is $f\left(\frac{3\pi}{4}ight) = 1$.Next,the left-hand limit is $\lim_{x 	o \frac{3\pi}{4}^-} f(x) = 1$.Then,the right-hand limit is $\lim_{x 	o \frac{3\pi}{4}^+} f(x) = \lim_{h 	o 0} 2\sin \left(\frac{2}{9} \left(\frac{3\pi}{4} + hight)ight) = 2\sin \left(\frac{2}{9} \cdot \frac{3\pi}{4}ight) = 2\sin \left(\frac{\pi}{6}ight) = 2 \cdot \frac{1}{2} = 1$.Since $\lim_{x 	o \frac{3\pi}{4}^-} f(x) = \lim_{x 	o \frac{3\pi}{4}^+} f(x) = f\left(\frac{3\pi}{4}ight) = 1$,the function $f(x)$ is continuous at $x = \frac{3\pi}{4}$.

If $f(x) = \begin{cases} 1, & 0 < x \le \frac{3\pi}{4} \\ 2\sin \frac{2}{9}x, & \frac{3\pi}{4} < x < \pi \end{cases}$,then

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