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
 A: 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
$$
\frac{(x-\sin x)\sin x}{(1-\cos x)^2} = \frac{\left(\frac{x^3}{6}+o(x^3)\right)(x+o(x))}{\left(\frac{x^2}{2}+o(x^2)\right)^2}
\operatorname*{\sim}_{x\to 0} \frac{\frac{x^3}{6}\cdot x }{\frac{x^4}{4}} = \frac{2}{3}.
$$
A: L'Hospital's rule works if you stick with it, I had to apply it 3 times.
$$\lim \limits_{x \to 0} \frac{(x-sinx)(1+cosx)}{(1-cosx)(sinx)}$$
$$=2\lim \limits_{x \to 0} \frac{(x-sinx)}{(sinx - \frac 12 sin(2x))} \to 2\lim \limits_{x \to 0} \frac{(1-\cos x)}{cos(x)-cos(2x)} $$
$$\to 2\lim \limits_{x \to 0} \frac{\sin x}{-sin(x)+2sin(2x)} $$
$$\to 2\lim \limits_{x \to 0} \frac{cos x}{-cos(x)+4cos(2x)} =\frac 23 $$
A: 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}
A: 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*}
