Answer
$-\dfrac {1}{2}=-0.5$
Work Step by Step
$\lim _{x\rightarrow 0}\dfrac {\cos x-1}{\sin ^{2}x}=\dfrac {\cos x-1}{1-\cos ^{2}x}=\dfrac {-\left( 1-\cos x\right) }{\left( 1-\cos x\right) \left( 1+\cos x\right) }=-\dfrac {1}{1+\cos x}=-\dfrac {1}{1+\cos 0}=-\dfrac {1}{2}=-0.5$
