Answer
The value of the limit is $0$.
Work Step by Step
The area of the semicircle with radius $r=\frac{PQ}{2}$ is:
$$A(\theta)=\frac{\pi \cdot r^{2}}{2}=\frac{\pi \cdot PQ^{2}}{8}$$
Using the law of cosines in the triangle it follows:
$$PQ^{2}=10^{2}+10^{2}-2\cdot 10 \cdot 10 \cdot \cos(\theta)$$
$$PQ^{2}=100+100-2\cdot 100 \cdot \cos(\theta)$$
$$PQ^{2}=200-200 \cdot \cos(\theta)$$
$$PQ^{2}=200(1-\cos(\theta))$$
So:
$$A(\theta)=\frac{\pi \cdot PQ^{2}}{8}=\frac{200\pi (1-\cos(\theta))}{8}$$
The area of the triangle is:
$$B(\theta)=\frac{1}{2} \cdot 10 \cdot 10 \cdot \sin(\theta)=50\sin(\theta)$$
So the limit is:
$$\lim\limits_{\theta \to 0^{+}}\frac{\frac{200\pi (1-\cos(\theta))}{8}}{50\sin(\theta)}$$
$$\lim\limits_{\theta \to 0^{+}}\frac{200\pi (1-\cos(\theta))}{400\sin(\theta)}$$
$$\lim\limits_{\theta \to 0^{+}}\frac{\pi (1-\cos(\theta))}{2\sin(\theta)}$$
The limit has an indeterminate form type $\frac{0}{0}$.
$$\lim\limits_{\theta \to 0^{+}}\frac{\pi (1-\cos(\theta))\theta}{2\theta\sin(\theta)}$$
$$\lim\limits_{\theta \to 0^{+}}\frac{\pi}{2} \cdot \frac{1-\cos(\theta)}{\theta}\cdot \frac{\theta}{\sin(\theta)}$$
$$\frac{\pi}{2} \cdot \lim\limits_{\theta \to 0^{+}}\frac{1-\cos(\theta)}{\theta}\cdot \lim\limits_{\theta \to 0^{+}}\frac{\theta}{\sin(\theta)}$$
We know that those limits are usual:
$$\lim\limits_{\theta \to 0^{+}}\frac{1-\cos(\theta)}{\theta}=0~~\text{and}~~\lim\limits_{\theta \to 0^{+}}\frac{\theta}{\sin(\theta)}=1$$
So:
$$\frac{\pi}{2} \cdot 0\cdot 1=0$$
The value of the limit is $0$.
