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

If you have seen what they are, and know how to use them, one easy and systematic way is to use Taylor expansions. Recalling that when $x\to0$, one has
$$
\sin x = x - \frac{x^3}{6} + o(x^3)
$$
and
$$
\cos x = 1 - \frac{x^2}{2} + o(x^2)
$$
you obtain
$$
\frac{(x-\sin x)\sin x}{(1-\cos x)^2} = \frac{\left(\frac{x^3}{6}+o(x^3)\right)(x+o(x))}{\left(\frac{x^2}{2}+o(x^2)\right)^2}
\operatorname*{\sim}_{x\to 0} \frac{\frac{x^3}{6}\cdot x }{\frac{x^4}{4}} = \frac{2}{3}.
$$
