$f(x) = \frac{k \cos x}{\pi - 2x}$ for $x \ne \frac{\pi}{2}$, and $f\left(\frac{\pi}{2}\right) = 5$. For which $k$ is $f$ continuous at $x = \frac{\pi}{2}$?

Explanation

1. 01

   Put $x = \frac{\pi}{2} - h$: $\cos x = \sin h$ and $\pi - 2x = 2h$, so $\frac{k \cos x}{\pi - 2x} = \frac{k \sin h}{2h} \to \frac{k}{2}$.

2. 02

   Continuity needs $\frac{k}{2} = 5$, so $k = 10$.