The Question
An arc $PQ$ of a circle subtends a central angle $\theta$ as in the figure. Let $A(\theta)$ be the area between the chord $PQ$ and the arc $PQ$. Let $B(\theta)$ be the area between the tangent lines $PR$, $QR$, and the arc.
Find:
$$\lim_{\theta\to0^+}{\frac{A(\theta)}{B(\theta)}}$$
What I've done
I have been able to show that
$$\lim_{\theta\to0^+}{\frac{A(\theta)}{B(\theta)}}=\frac{1}{\lim_{\theta\to0^+}{\frac{(1-\cos\theta)\sqrt{2(1-\cos\theta)-\sin^2\theta}}{(\theta-\sin\theta)\sin\theta}}-1}$$
and using a computer algebra system, I got the answer 2. But I do not know how the computer got the answer and searching the web did not help at all. Can someone that knows give an answer?
Thanks in advance.
Best Answer
Here are my calculations, giving a different formula but the same limit $2$.
Let us assume, without loss of generality, that the radius of the circle is $1$.
Let $\alpha = \frac12 \theta$:
$$\begin{cases}\text{Using area of triangle } OPQ: & A(\theta)&=&\alpha - \sin \alpha \cos \alpha\\ \text{Using area of triangle } OPR: &A(\theta)+B(\theta)&=&\tan \alpha - \sin \alpha \cos \alpha \end{cases}$$ giving
$$B(\theta)=\tan \alpha - \alpha$$
Therefore :
$$\lim_{\theta \to 0^+}\frac{A(\theta)}{B(\theta)}=\lim_{\alpha \to 0^+}\frac{\alpha - \sin \alpha \cos \alpha}{\tan \alpha - \alpha}=\lim_{\alpha \to 0^+}\frac{\alpha -(\alpha - \tfrac16 \alpha^3+\cdots)(1-\tfrac12\alpha^2+\cdots)}{(\alpha+\tfrac13\alpha^3-\cdots) - \alpha}$$
(using Taylor expansions) ; as the numerator is $\approx \frac23\alpha^3$, the denominator is $\approx \frac13\alpha^3$, the limit is equal to $2$.