Limit of a function using Taylor series 
How can I verify that
  $\lim_{x \to 0}  \dfrac{\cos{2x} - \sqrt{1-4x^2}}{2x \sin x^3} = \frac{4}{3}$?

I've tried with Taylor so that:
$2x\sin x^3  \sim 4x^4 - \frac{2}{3}x^{10}$
$\cos2x \sim 1- 2x^2$
$\sqrt{1-4x^2} \sim 1-2x^2-2x^4$
But it keeps me giving the wrong result, so the question is, am I using wrong the Taylor series? 
When approximating the function and plugging them in the limit should them all be of the same order?
A well-detailed explanation is more than welcome.

Hint. First of all $2x\sin(x)^3=2x^4+o(x^4)$. Moroever, here you need a longer expansion for $\cos(2x)$:
$$\cos(2x)=1-\frac{(2x)^2}{2}+\frac{(2x)^4}{4!}+o(x^4)=1-2x^2+\frac{2x^4}{3}+o(x^4).$$
Hence
$$\frac{\cos(2x) - \sqrt{1-4x^2}}{2x \sin x^3}=
\frac{1-2x^2+\frac{2x^4}{3} - (1-2x^2-2x^4)+o(x^4)}{2x^4+o(x^4)}
$$
Do you mind to give another try?
P.S. I suggest the use of the little-o notation. It will help you to understand whether a Taylor expansion is sufficient for the purpose.
