Finding $\lim_{x\to 0} \frac {2\sin x-\sin 2x}{x-\sin x}$ geometrically

While looking at this question, I noticed an interesting geometric interpretation of the limit the OP was trying to evaluate. His limit came to twice the value of the limit

$$\lim_{x\to 0}\frac{\sin x (1-\cos(x))}{x-\sin x}$$

which can be interpreted as the limit of the ratio of the area of $\triangle DBC$ and the area bounded between the line $CB$ and the arc $CB$. (See the image below). I know that the limit $\lim_{x\to 0} \frac{\sin x}{x} = 1$ is often derived through a geometric method. Can this the limit I've given be evaluated geometrically?

• This is an exercise in Stewart. It can of course be done by L'Hopital's rule, but L'H's rule, although it may find the answer very efficiently, often doesn't give much if any insight. – Michael Hardy Jul 7 '15 at 5:38
• If we let $f(x) = x - \sin x$, then the original limit is $\lim_{x \to 0} [f(2x)/f(x) - 2]$. So in order to calculate the limit, it is enough to establish that $f(x) \sim ax^3$ for some nonzero $a$. We don't need to know that $a = 1/6$. I'm not sure whether this makes things any easier. – Keith Jul 7 '15 at 6:06

Not sure how convincing this is, but consider: That point in the middle of trangle $DBC$ is its centroid, so that all three subtriangles have the same area. Does the area of the upper right subtriangle approximate the area of the arc sliver to a high enough degree as the angle approaches $0$? If so, the ratio is of course $3$.
• Visually it looks convincing but this is not as rigorous as the proof of $(\sin x)/x \to 1$ as $x \to 0$ given at math.stackexchange.com/a/75151/72031 I believe it is impossible to find a geometric rigorous proof of this limit as it is equivalent to the limit $(x - \sin x)/x^{3} \to 1/6$ as $x \to 0$ and this is one is notoriously difficult to prove without using derivatives. – Paramanand Singh Jul 7 '15 at 11:45
I think this limit is impossible to evaluate geometrically, at least using the squeeze theorem. First, note that the area of $∆ACB=sinx/2$ because the area of a triangle is $bh/2$ with $b=1$ and $sinx=h$ in this case. Also, $∆ADC=sin(x)cos(x)/2$ and the area of the sector is $x/2$. Thus, to use the Squeeze Theorem, we would need this inequality:
$sinxcosx/2 < sinx/2 < x/2$