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

First, write
\begin{equation*}
\frac{(x-\sin x)\sin x}{(1-\cos x)^{2}}=\left( \frac{x-\sin x}{x^{3}}\right)
\left( \frac{\cos (x)-1}{x^{2}}\right) ^{-2}\left( \frac{\sin x}{x}\right) 
\end{equation*}
Next, using the standard limits
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}} &=&\frac{1}{6} \\
\lim_{x\rightarrow 0}\frac{\cos x-1}{x^{2}} &=&-\frac{1}{2} \\
\lim_{x\rightarrow 0}\frac{\sin x}{x} &=&1
\end{eqnarray*}
It follows that
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{(x-\sin x)\sin x}{(1-\cos x)^{2}}
&=&\lim_{x\rightarrow 0}\left( \frac{x-\sin x}{x^{3}}\right) \left( \frac{%
\cos (x)-1}{x^{2}}\right) ^{-2}\left( \frac{\sin x}{x}\right)  \\
&=&\lim_{x\rightarrow 0}\left( \frac{x-\sin x}{x^{3}}\right) \cdot \left(
\lim_{x\rightarrow 0}\frac{\cos (x)-1}{x^{2}}\right) ^{-2}\cdot
\lim_{x\rightarrow 0}\left( \frac{\sin x}{x}\right)  \\
&=&\left( \frac{1}{6}\right) \cdot \left( -\frac{1}{2}\right) ^{-2}\cdot 1=%
\frac{2}{3}.\ \ \ \ \blacksquare 
\end{eqnarray*}
