# Intuition for a limit found using L'Hôpital (geometry)

$$OPR$$ 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 $$\theta \rightarrow 0$$?

• 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). – egreg Feb 13 at 10:07

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.
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 $$\sin x \sim x - \frac{x^3}{6}$$.