Answer
$0$
Work Step by Step
The area $A(\theta)$ is the area of the sector with angle $\theta$ and radius $QR$ minus the area of the triangle $\triangle POR$ it follows:
$$A(\theta)=\frac{1}{2}OR^{2} \theta-\frac{1}{2}PQ \cdot OR$$
From the diagram we have $OP=OR$ because they are radii of the arc and:
$$\sin(\theta)=\frac{PQ}{OP} \to \sin(\theta)=\frac{PQ}{OR} \to PQ=OR\sin(\theta) $$
So:
$$A(\theta)=\frac{1}{2}OR^{2} \theta-\frac{1}{2}OR\sin(\theta) \cdot OR$$
$$A(\theta)=\frac{1}{2}OR^{2} \theta-\frac{1}{2}OR^{2}\sin(\theta)$$
$$A(\theta)=\frac{1}{2}OR^{2}( \theta-\sin(\theta))$$
The area $B(\theta)$ is:
$$B(\theta)=\frac{1}{2}PQ \cdot QR \to B(\theta)=\frac{1}{2}OR\sin(\theta) \cdot QR $$
So the limit is:
$$ \lim _{ \theta ^{+} \to 0}\frac{OR( \theta-\sin(\theta))}{\sin(\theta) \cdot QR}$$
$$ \lim _{ \theta ^{+} \to 0}\frac{OR}{QR}\cdot \frac{( \theta-\sin(\theta))}{\sin(\theta)}$$
$$\frac{OR}{QR}\cdot \lim _{ \theta ^{+} \to 0}\frac{( \theta-\sin(\theta))}{\sin(\theta)}$$
The limit has an indeterminate form type $\frac{0}{0}$ so we apply the l'Hospital's rule it follows:
$$\frac{OR}{QR}\cdot\lim _{ \theta ^{+} \to 0} \frac{( \theta-\sin(\theta))'}{(\sin(\theta))'}$$
$$\frac{OR}{QR}\cdot\lim _{ \theta ^{+} \to 0} \frac{1-\cos(\theta)}{\cos(\theta)}$$
$$\frac{OR}{QR}\cdot \frac{1-\cos(0)}{\cos(0)}=0$$
