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Let $f(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{\pi}{2}\right]$. If $f(x)$ is continuous in $\left[0, \frac{\pi}{2}\right]$, then $f\left(\frac{\pi}{4}\right)$ is
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$-\frac{1}{2}$
$-\frac{1}{2}$
$f(x)=\frac{1-\tan x}{4 x-\pi} \Rightarrow \lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\tan x}{4 x-\pi}=-\frac{1}{2}$
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