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? 
 A: \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}
A: Use Taylor expansion:
$$
1-\cos x=1-1+\frac{x^2}{2}+o(x^2)=\frac{x^2}{2}+o(x^2)
$$
so you need to go to degree $2$ also in the numerator.
Use $\sin2x=2x+o(x^2)$ and
$$
\sqrt[3]{1+x}=1+\frac{x}{3}+o(x)
$$
so you have
$$
\lim_{x\to0}\frac{2x+o(x^2)-2x(1+x/3+o(x))}{x^2/2+o(x^2)}=
\lim_{x\to0}\frac{-2x^2/3+o(x^2)}{x^2/2+o(x^2)}=-\frac{4}{3}
$$
A: Using L'Hôpital's Rule
$$\lim_{x\to 0}\frac{\sin2x-2x(1+x)^{1/3}}{1-\cos x}$$
Try x = 0
$$\lim_{x\to 0}\frac{\sin0-2*0(1)^{1/3}}{1-\cos 0} = \frac{0}{0}$$
Derive f(x) and g(x) and set x = 0
$$\frac{f'(x)}{g'(x)}=\lim_{x\to 0}\frac{2\cos2x-\frac{2(4x+3)}{3(x+1)^{2/3}}}{-\sin x}= \frac{0}{0}$$
Derive again and set x = 0
$$\frac{f''(x)}{g''(x)}=\lim_{x\to 0}\frac{-4\sin2x-\frac{4(2x+3)}{9(x+1)^{5/3}}}{\cos x}= -\frac{4}{3}$$
