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.

You are just replacing $\sin x$ by $x$ and $\cos \sqrt{x^3}$ with $1-(x^3/2)$. Both these replacements are wrong for the very simple reason that $\sin x\neq x$ and $\cos \sqrt{x^3}\neq 1-(x^3/2)$ unless $x=0$.
The right approach is to use the standard limits $$\lim_{x\to 0}\frac{\sin x} {x} =1,\,\lim_{x\to 0}\frac{1-\cos x} {x^2}=\frac{1}{2}\tag{1}$$ Using above limit and L'Hospital's Rule it can be easily proved that $$\lim_{x\to 0}\frac{x-\sin x} {x^3}=\frac {1}{6}\tag{2}$$ Another limit which is needed here is $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}\tag{3}$$ We can evaluate the limit in question using limits $(1), (2)$ and $(3)$ as follows
\begin{align}
L&=\lim_{x\to 0}\frac{2x-\sin x\sqrt[3]{8-x^2}}{1-\cos\sqrt{x^3}}\notag\\
&= \lim_{x\to 0}\frac{2x-\sin x\sqrt[3]{8-x^2}}{x^3}\cdot\frac{x^3}{1-\cos\sqrt{x^3}}\notag\\
&= 2\lim_{x\to 0}\frac{2x-2\sin x+\sin x(2-\sqrt[3]{8-x^2}) } {x^3}\text{ (using (1))}\notag\\
&=2\left(\frac{1}{3}+\lim_{x\to 0}\frac{\sin x} {x} \cdot\frac{2-\sqrt[3]{8-x^2}}{x^2}\right)\text{ (using (2))}\notag\\
&= \frac{2}{3}+2\lim_{t\to 8}\frac{8^{1/3}-t^{1/3}}{8-t} \text{ (putting }t=8-x^2) \notag\\
&= \frac{2}{3}+2\cdot\frac{1}{3}\cdot 8^{-2/3}\text{ (using (3))}\notag\\
&=\frac{5}{6}\notag
\end{align} 
