If the function f is continuous at x = 1, whe | vedclass

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(C) Since $f$ is continuous at $x = 1$, $f(1) = \lim_{x 	o 1} f(x)$.Let $x = 1 + h$. As $x 	o 1$, $h 	o 0$.$f(1) = \lim_{h 	o 0} \frac{1 + \cos(\p

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$\frac{\pi}{2}$

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(C) Since $f$ is continuous at $x = 1$, $f(1) = \lim_{x 	o 1} f(x)$.Let $x = 1 + h$. As $x 	o 1$, $h 	o 0$.$f(1) = \lim_{h 	o 0} \frac{1 + \cos(\pi(1+h))}{\pi(1-(1+h))^2}$.Using $\cos(\pi + 	heta) = -\cos(	heta)$, we get $\cos(\pi + \pi h) = -\cos(\pi h)$.$f(1) = \lim_{h 	o 0} \frac{1 - \cos(\pi h)}{\pi h^2}$.Using the identity $1 - \cos(	heta) = 2\sin^2(	heta/2)$, we get $1 - \cos(\pi h) = 2\sin^2(\frac{\pi h}{2})$.$f(1) = \lim_{h 	o 0} \frac{2\sin^2(\frac{\pi h}{2})}{\pi h^2} = \lim_{h 	o 0} \frac{2}{\pi} \left( \frac{\sin(\frac{\pi h}{2})}{\frac{\pi h}{2} \cdot \frac{2}{\pi}} ight)^2 = \lim_{h 	o 0} \frac{2}{\pi} \cdot \frac{\pi^2}{4} \cdot \left( \frac{\sin(\frac{\pi h}{2})}{\frac{\pi h}{2}} ight)^2$.Since $\lim_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$, we get $f(1) = \frac{2}{\pi} \cdot \frac{\pi^2}{4} \cdot 1^2 = \frac{\pi}{2}$.

If the function $f$ is continuous at $x = 1$, where $f(x) = \frac{1 + \cos(\pi x)}{\pi(1-x)^2}$ for $x \neq 1$, then the value of $f(1)$ is....

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