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?
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 $3$.
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:

$sinxcosx/2 &lt; sinx/2 &lt; x/2$

The problem is that this inequality only holds true in quadrant 1 where everything is positive, but in order to take the limit as x goes to 0, we need to evaluate the limit from quadrant 4 because we take the limit from left to right. But because sinx is negative in quad 4, the inequality no longer holds true. Obviously, we would need to majorly rewrite the inequality before we can even take the limit, but I strongly suspect that we'd still have the same problem. In fact, I suspect that no matter how we set up the inequality, this problem still remains. So I doubt you can evaluate this limit geometrically, using basic methods anyway.