Let f(x) = begin{cases} frac{1 + cos 2pi x}{1 | vedclass

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(A) For $f(x)$ to be continuous at $x = \frac{1}{2}$,we must have $\lim_{x 	o \frac{1}{2}^-} f(x) = \lim_{x 	o \frac{1}{2}^+} f(x) = f(\frac{1}{2})

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$p \in R - \{4\}$

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(A) For $f(x)$ to be continuous at $x = \frac{1}{2}$,we must have $\lim_{x 	o \frac{1}{2}^-} f(x) = \lim_{x 	o \frac{1}{2}^+} f(x) = f(\frac{1}{2}) = p$.First,calculate the left-hand limit:$\lim_{x 	o \frac{1}{2}^-} \frac{1 + \cos 2\pi x}{1 - \sin \pi x} = \lim_{x 	o \frac{1}{2}^-} \frac{2 \cos^2 \pi x}{1 - \sin \pi x} = \lim_{x 	o \frac{1}{2}^-} \frac{2(1 - \sin^2 \pi x)}{1 - \sin \pi x} = \lim_{x 	o \frac{1}{2}^-} 2(1 + \sin \pi x) = 2(1 + 1) = 4$.Next,calculate the right-hand limit:Let $t = \sqrt{2x - 1}$. As $x 	o \frac{1}{2}^+$,$t 	o 0^+$.$\lim_{t 	o 0^+} \frac{t}{\sqrt{4 + t} - 2} = \lim_{t 	o 0^+} \frac{t(\sqrt{4 + t} + 2)}{(4 + t) - 4} = \lim_{t 	o 0^+} (\sqrt{4 + t} + 2) = \sqrt{4} + 2 = 4$.Since $\lim_{x 	o \frac{1}{2}^-} f(x) = \lim_{x 	o \frac{1}{2}^+} f(x) = 4$,the function is continuous at $x = \frac{1}{2}$ if $p = 4$.Therefore,$f(x)$ is discontinuous at $x = \frac{1}{2}$ if $p 
eq 4$,which means $p \in R - \{4\}$.

Let $f(x) = \begin{cases} \frac{1 + \cos 2\pi x}{1 - \sin \pi x}, & x < \frac{1}{2} \\ p, & x = \frac{1}{2} \\ \frac{\sqrt{2x - 1}}{\sqrt{4 + \sqrt{2x - 1}} - 2}, & x > \frac{1}{2} \end{cases}$. If $f(x)$ is discontinuous at $x = \frac{1}{2}$,then:

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