Given, \(\lim\limits_{\text x \to\pi/2}\cfrac{\sqrt{2-sin\,\text x}-1}{(\frac \pi 2-\text x)^2}
\)
Now, rationalize the Numerator, we get,
Hence, \(\lim\limits_{\text x \to\pi/2}\cfrac{\sqrt{2-sin\,\text x}-1}{(\frac \pi 2-\text x)^2}
\) = \(\cfrac14\)