How do I solve $\lim_{x\to 0}\frac{\sin2x-2x(1+x)^{1/3}}{1-\cos x}$? I'm trying to get my head around this equation,
$$\lim_{x\to 0}\frac{\sin2x-2x(1+x)^{1/3}}{1-\cos x}$$
but nothing I do seems to make it any more clearer. Do any one know how to do it? 

\begin{align}
\lim_{x\to 0}\frac{\sin2x-2x(1+x)^{1/3}}{1-\cos x}
&=\lim_{x\to 0}\frac{\sin2x-2x(1+x)^{1/3}}{x^2}\cdot\frac{x^2}{1-\cos x}
\\
&=\lim_{x\to 0}\frac{\sin2x-2x(1+x)^{1/3}}{x^2}\cdot\lim_{x\to 0}\frac{x^2}{1-\cos x}\\
&=\lim_{x\to 0}\frac{\sin2x-2x(1+x)^{1/3}}{x^2}\cdot2
\\
&=2\lim_{x\to 0}\frac{2x-\frac{8x^3}{3!}+O(x^5)-2x-\frac23x^2-O(x^3)}{x^2}\\
&=2\lim_{x\to 0}\frac{-\frac{2x^2}{3}+O(x^5)-O(x^3)}{x^2}
\\
&=-\frac{4}3
\end{align}
