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$OPR$ is a sector with central angle $\theta$. $A(\theta)$ is the area of the segment bounded by the line $PR$ and the arc $PR$ and $B(\theta)$ is the area of the triangle $PQR$.

The ratio $$\frac{A(\theta)}{B(\theta)} = \frac{r^2(\theta-\sin\theta)}{2\frac{r^2(1-\cos\theta)\sin\theta}{2}}$$

If I use L'Hôpital to find $\lim_{\theta \rightarrow 0} \frac{A(\theta)}{B(\theta)}$, then the answer is $\frac{1}{3}$.

I was wondering if there is any intuition for this limit and if this should be the answer you 'expect'... I originally thought it would be $0$ and also not sure why this is wrong.

As a smaller note, most books with this example write $\theta \rightarrow 0^+$, but is it OK to still write $\theta \rightarrow 0$?

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    $\begingroup$ About notation: it depends on conventions. Since the function is only defined for $0<\theta<\pi$, using $\theta\to0$ is perfectly acceptable (but opinions may differ). $\endgroup$ – egreg Feb 13 at 10:07
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Doing the same as @Martín-Blas Pérez Pinilla but using more terms, you can have not only the limit but also a quite good approximation using $$\frac{\theta - \sin(\theta)}{(1 - \cos(\theta))\sin(\theta)} =\frac{\frac{t^3}{6}-\frac{t^5}{120}+O\left(t^7\right) }{\frac{t^3}{2}-\frac{t^5}{8}+O\left(t^7\right) }=\frac{1}{3}+\frac{t^2}{15}+O\left(t^4\right)$$

Just for the fun, make $\theta=\frac \pi 6$. The exact answer is $\frac{2 (\pi -3)}{3 \left(2-\sqrt{3}\right)}\approx 0.3523$ while the above approximation would give $\frac{1}{3}+\frac{\pi ^2}{540}\approx 0.3516$.

Concerning the notations, if $\lim_{\theta\to 0^+}=\lim_{\theta\to 0^-}$, in my opinion, it does not matter and $\lim_{\theta\to 0}$ is correct.

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Not terribly intuitive, but using Taylor in the numerator and a trigonometric formula in the denominator: $$\frac{(\theta - \sin\theta)}{(1 - \cos\theta)\sin\theta} = \frac{\theta^3/6 + o(\theta^3)}{2\sin^2(\theta/2)\sin\theta},$$ the equivalence $\sin\theta\approx\theta$ for $\theta$ small gives easily the answer.

Update: the approximation $\sin\theta \sim \theta - \theta^3/6$ has a geometric proof: Is there a geometric method to show $\sin x \sim x - \frac{x^3}{6}$.

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