$\lim\limits_{x\to 0}\dfrac{1-\sqrt{\cos x}}{\sin^2x}$
$=\lim\limits_{x\to 0}\dfrac{1-\sqrt{\cos x}}{1-\cos^2x}$
$=\lim\limits_{x\to 0}\dfrac{1-\sqrt{\cos x}}{(1+\cos x)(1-\sqrt{\cos x})(1+\sqrt{\cos x})}$
$=\lim\limits_{x\to 0}\dfrac{1}{(1+\cos x)(1+\sqrt{cos x})}$
$=\dfrac{1}{(1+1)(1+1)}$
$=\dfrac{1}{4}$