A Limit of Complex Trig Please help with a messy trig. I only want the process of evaluating the limits! Appreciate!
$$\lim_{x \to 0} \frac{(x-\sin x)\sin x}{(1-\cos x)^2}$$
or the alternative form 
$$\lim_{x \to 0} \frac{(x-\sin x)(1+\cos x)}{(1-\cos x)(\sin x)}$$
Wolframalpha gave $2/3$
The original question(with image) is asking the area ratio. $\frac{ABD}{ADBC}$
colored segment ABD over the triangle ABC minus segment ABD

Best is to use L'Hospital combined with standard limits (and in my opinion using just L'Hospital is a bad idea). We use the standard limit $$\lim_{x \to 0}\frac{\sin x}{x} = 1$$ and almost standard limit $$\lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \lim_{x \to 0}\frac{(1 - \cos x)(1 + \cos x)}{x^{2}(1 + \cos x)} = \frac{1}{2}\lim_{x \to 0}\frac{\sin^{2}x}{x^{2}} = \frac{1}{2}$$ Using these we can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{(x - \sin x)\sin x}{(1 - \cos x)^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{(x - \sin x)\sin x}{\dfrac{(1 - \cos x)^{2}}{x^{4}}\cdot x^{4}}\notag\\
&= \lim_{x \to 0}\dfrac{(x - \sin x)\sin x}{\dfrac{1}{4}\cdot x^{4}}\notag\\
&= 4\lim_{x \to 0}\frac{\sin x}{x}\cdot\frac{x - \sin x}{x^{3}}\notag\\
&= 4\lim_{x \to 0}\frac{x - \sin x}{x^{3}}\notag\\
&= 4\lim_{x \to 0}\frac{1 - \cos x}{3x^{2}}\text{ (via L'Hospital's Rule)}\notag\\
&= \frac{4}{3}\cdot\frac{1}{2} = \frac{2}{3}\notag
\end{align}
