I'm not going to work through all of the details of the area calculations. It's just elementary trigonometry.
The area of POR is $\displaystyle \frac{r^2}{2}\sin\theta$. The area of the whole sector is $\displaystyle \pi r^2\cdot\frac{\theta}{2\pi}=\frac{r^2\theta}{2}$. So the area of $\displaystyle A(\theta)$ is $\displaystyle \frac{r^2}{2}(\theta-\sin\theta)$.
Furthermore, the area of POQ is $\displaystyle \frac{r^2}{2}\cos\theta\sin\theta=\frac{r^2}{2}\cd ot\frac{\sin2\theta}{2}$ so $\displaystyle B(\theta)=\frac{r^2}{2}(\sin\theta-\frac{\sin2\theta}{2})$
$\displaystyle \lim_{\theta\to0^+}\frac{A(\theta)}{B(\theta)}=\fr ac{\frac{r^2}{2}(\theta-\sin\theta)}{\frac{r^2}{2}(\sin\theta-\frac{\sin2\theta}{2})}=$ $\displaystyle \lim_{\theta\to0^+}\frac{\theta-\sin\theta}{\sin\theta-\frac{\sin2\theta}{2}}$
Use L'Hopital's Rule thrice to get that the limit equals $\displaystyle \frac{1}{3}$.