Let f(x) = frac{1 - tan x}{4x - pi }, x ne fr | vedclass

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(C) For $f(x)$ to be continuous at $x = \frac{\pi }{4}$,we must have $f(\frac{\pi }{4}) = \lim_{x 	o \frac{\pi }{4}} f(x)$.Evaluating the limit: $\li

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$-\frac{1}{2}$

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(C) For $f(x)$ to be continuous at $x = \frac{\pi }{4}$,we must have $f(\frac{\pi }{4}) = \lim_{x 	o \frac{\pi }{4}} f(x)$.Evaluating the limit: $\lim_{x 	o \frac{\pi }{4}} \frac{1 - 	an x}{4x - \pi }$.Since this is a $\frac{0}{0}$ indeterminate form,we apply $L$'$H$ôpital's rule:$\lim_{x 	o \frac{\pi }{4}} \frac{\frac{d}{dx}(1 - 	an x)}{\frac{d}{dx}(4x - \pi )} = \lim_{x 	o \frac{\pi }{4}} \frac{-\sec^2 x}{4}$.Substituting $x = \frac{\pi }{4}$:$\frac{-\sec^2(\frac{\pi }{4})}{4} = \frac{-(\sqrt{2})^2}{4} = \frac{-2}{4} = -\frac{1}{2}$.Therefore,$f(\frac{\pi }{4}) = -\frac{1}{2}$.

Let $f(x) = \frac{1 - \tan x}{4x - \pi }, x \ne \frac{\pi }{4}, x \in [0, \frac{\pi }{2}]$. If $f(x)$ is continuous in $[0, \frac{\pi }{2}]$,then $f(\frac{\pi }{4})$ is

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