<s>[INST] 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
 [/INST] 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
$$
\frac{(x-\sin x)\sin x}{(1-\cos x)^2} = \frac{\left(\frac{x^3}{6}+o(x^3)\right)(x+o(x))}{\left(\frac{x^2}{2}+o(x^2)\right)^2}
\operatorname*{\sim}_{x\to 0} \frac{\frac{x^3}{6}\cdot x }{\frac{x^4}{4}} = \frac{2}{3}.
$$
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