If f(x) = frac{1 - sin x}{log(1 + pi^2 - 4pi | vedclass

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(B) Since $f(x)$ is continuous at $x = \frac{\pi}{2}$,we have $f\left(\frac{\pi}{2}ight) = \lim_{x 	o \frac{\pi}{2}} f(x)$.Let $t = x - \frac{\pi}{

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$\frac{1}{8}$

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(B) Since $f(x)$ is continuous at $x = \frac{\pi}{2}$,we have $f\left(\frac{\pi}{2}ight) = \lim_{x 	o \frac{\pi}{2}} f(x)$.Let $t = x - \frac{\pi}{2}$. As $x 	o \frac{\pi}{2}$,$t 	o 0$.Then $x = t + \frac{\pi}{2}$.The denominator becomes $\log(1 + \pi^2 - 4\pi(t + \frac{\pi}{2}) + 4(t + \frac{\pi}{2})^2) = \log(1 + \pi^2 - 4\pi t - 2\pi^2 + 4(t^2 + \pi t + \frac{\pi^2}{4})) = \log(1 + 4t^2)$.The numerator becomes $1 - \sin(t + \frac{\pi}{2}) = 1 - \cos t$.So,$f\left(\frac{\pi}{2}ight) = \lim_{t 	o 0} \frac{1 - \cos t}{\log(1 + 4t^2)}$.Using the standard limits $\lim_{t 	o 0} \frac{1 - \cos t}{t^2} = \frac{1}{2}$ and $\lim_{u 	o 0} \frac{\log(1 + u)}{u} = 1$:$f\left(\frac{\pi}{2}ight) = \lim_{t 	o 0} \frac{(1 - \cos t)/t^2}{\log(1 + 4t^2)/(4t^2) 	imes 4} = \frac{1/2}{1 	imes 4} = \frac{1}{8}$.

If $f(x) = \frac{1 - \sin x}{\log(1 + \pi^2 - 4\pi x + 4x^2)}$ is continuous at $x = \frac{\pi}{2}$,then $f\left(\frac{\pi}{2}\right) = $

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