If $f(x) = \frac{k \cos x}{\pi - 2x}$ for $x \neq \frac{\pi}{2}$ and $f\left(\frac{\pi}{2}\right) = 3$ is continuous at $x = \frac{\pi}{2}$, then the value of $k$ is:
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Apply L'Hopital's Rule for $\frac{0}{0}$ limits: differentiate numerator and denominator with respect to $x$: $\lim_{x \to \pi/2} \frac{-k\sin x}{-2} = \frac{k}{2} = 3 \implies k = 6$. It's super fast!