This math has $\log x$ as power of the $\sin x$ trigonometric function.
$$\lim _{x\to\pi/2}\left( \sin x\right)^{\log x}$$
2026-04-13 06:14:44.1776060884
Evaluating $\lim _{x\to\pi/2}\left( \sin x\right)^{\log x}$
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The function is continuous at $\frac{\pi}{2}$.
Hence,
$$\lim_{x\to \frac{\pi}{2}}\left(\sin x\right)^{\log x} =1^{\log \frac{\pi}{2}} = 1 $$