\(\lim\limits_{\text x \to0}\cfrac{\sqrt2-\sqrt{1+cos\,\text x}}{sin^2\text x} \)
Rationalize the numerator, we get
\(\Rightarrow\lim\limits_{\text x \to0}\cfrac{\sqrt2-\sqrt{1+cos\,\text x}}{sin^2\text x} \)
Hence, \(\lim\limits_{\text x \to0}\cfrac{\sqrt2-\sqrt{1+cos\,\text x}}{sin^2\text x} \) = \(\cfrac12\)