Answer
$\dfrac {1}{4}=0.25$
Work Step by Step
$$\lim _{\theta \rightarrow 0}\dfrac {\dfrac {1}{2+\sin \theta }-\dfrac {1}{2}}{\sin \theta }=\dfrac {\dfrac {2+\sin \theta -2}{2\times \left( 2+\sin \theta \right) }}{\sin \theta }=\dfrac {\dfrac {\sin \theta }{2\times \left( 2+\sin \theta \right) }}{\sin \theta }=\dfrac {1}{2\times \left( 2+\sin \theta \right) }=\dfrac {1}{2\times \left( 2+0\right) }=\dfrac {1}{4}=0.25$$