If f(x) = frac{1+cos pi x}{pi(1-x)^2} for x n | vedclass

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(C) For $f(x)$ to be continuous at $x=1$,$f(1) = \lim_{x 	o 1} f(x)$.Let $1-x = h$. As $x 	o 1$,$h 	o 0$.$f(1) = \lim_{h 	o 0} \frac{1+\cos(\pi(1-

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$\frac{\pi}{2}$

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(C) For $f(x)$ to be continuous at $x=1$,$f(1) = \lim_{x 	o 1} f(x)$.Let $1-x = h$. As $x 	o 1$,$h 	o 0$.$f(1) = \lim_{h 	o 0} \frac{1+\cos(\pi(1-h))}{\pi h^2}$$f(1) = \lim_{h 	o 0} \frac{1+\cos(\pi - \pi h)}{\pi h^2}$Since $\cos(\pi - 	heta) = -\cos 	heta$,we have:$f(1) = \lim_{h 	o 0} \frac{1-\cos(\pi h)}{\pi h^2}$Using the identity $1-\cos 	heta = 2\sin^2(\frac{	heta}{2})$:$f(1) = \lim_{h 	o 0} \frac{2\sin^2(\frac{\pi h}{2})}{\pi h^2}$$f(1) = \frac{2}{\pi} \lim_{h 	o 0} \left( \frac{\sin(\frac{\pi h}{2})}{h} ight)^2$Multiply and divide by $(\frac{\pi}{2})^2$:$f(1) = \frac{2}{\pi} \cdot (\frac{\pi}{2})^2 \lim_{h 	o 0} \left( \frac{\sin(\frac{\pi h}{2})}{\frac{\pi h}{2}} ight)^2$$f(1) = \frac{2}{\pi} \cdot \frac{\pi^2}{4} \cdot (1)^2 = \frac{\pi}{2}$.

If $f(x) = \frac{1+\cos \pi x}{\pi(1-x)^2}$ for $x \neq 1$ is continuous at $x=1$,then $f(1)$ is equal to

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