calculus - Simple limit with asymptotic approach. Where's the error?

Simply calculus question about a limit.

I don't understand why I'm wrong, I have to calculate

$$\lim_{x \rightarrow 0} \frac{2x - \sqrt[3]{8 - x^2}\sin x}{1 - \cos\sqrt{x^3}}$$

Using asymptotics, limits and De l'Hospital rule I would write these passages...

$$= \lim_{x \rightarrow 0} \frac{x \, (2 - \sqrt[3]{8 - x^2})}{x^3/2} = \lim_{x \rightarrow 0} \frac{\frac{2}{3}\frac{\sqrt[3]{8 - x^2}}{8-x^2}x}{x} = \frac{1}{6}$$

But the answers should be $\frac{5}{6}$. Thank you for your help.

Answer

The mistake lies at the beginning :

$$\lim_{x \rightarrow 0} \frac{2x - \sqrt[3]{8 - x^2}(x-\frac{x^3}{6})}{1 - \cos\sqrt{x^3}} =\frac56$$

$$\lim_{x \rightarrow 0} \frac{2x - \sqrt[3]{8 - x^2}\:x}{1 - \cos\sqrt{x^3}} =\frac16$$

At denominator $1-\cos(x^{3/2})$ is equivalent to $\frac12 x^3$. Thus one cannot neglect the $x^3$ terms in the numerator. So, the equivalent of $\sin(x)$ must not be $x$ but $x-\frac{x^3}{6}$ . This was the trap of the exercise.