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.