Can't solve following limit: $\lim_{x \to \infty}x \left( \sqrt[3]{5+8x^3} - 2x\right)$ Need to solve following problem:
$$\lim_{x \to \infty}x \left( \sqrt[3]{5+8x^3} - 2x\right) $$ I've tried to do something like this: $$\lim_{x \to \infty} x\left(\sqrt[3]{5+8x^3} - 2x\right) =\lim_{x \to \infty}x\left( \sqrt[3]{ \left( \frac {5}{x^3}+8 \right)x^3 } - 2x\right) = \lim_{x \to \infty} x\left( \left ( \sqrt[3] { \frac{5}{x^3}+8} \right)x - 2x\right) $$ 
It seems to be the right way, but I can't do my next step.
 A: Hint:
$$\sqrt[3]{5+8x^3} - 2x = \sqrt[3]{5+8x^3} - \sqrt[3]{8x^3} = \frac{\left(5+8x^3\right)-8x^3}{\left(\sqrt[3]{5+8x^3}\right)^2 + \sqrt[3]{5+8x^3} \cdot \sqrt[3]{8x^3} + \left(\sqrt[3]{8x^3}\right)^2}.$$
A: Note that 
$$2x\lt \sqrt[3]{5+8x^3}=2x\left(1+\frac{5}{8x^3}\right)^{1/3}\lt 2x\left(1+\frac{5}{24x^3}\right).$$
Then use Squeezing. 
A: Putting $n=\frac1h$
$$\lim_{x \to \infty}x \left( \sqrt[3]{5+8x^3} - 2x\right) $$
$$=\lim_{h\to0}\frac{\sqrt[3]{5h^3+8}-2}{h^2}$$
$$=\lim_{h\to0}\frac{5h^3+8-2^3}{h^2\{(5h^3+8)^{\frac23}+(5h^3+8)^{\frac13}+1\}}$$
$$=\lim_{h\to0}\frac{5h}{\{(5h^3+8)^{\frac23}+(5h^3+8)^{\frac13}+1\}}$$
$$=\cdots$$
A: $$\begin{aligned} \lim _{x\to \infty }\left(2x^2\:\left(\:\sqrt[3]{1+\frac{5}{8x^3}}\:-\:1\right)\right)
\\&\approx \lim _{x\to \infty }\left(2x^2\:\left(\:\frac{5}{8x^3}\cdot \frac{1}{3}\:\right)\right)
\\&=\lim _{x\to \infty }\left(\frac{5}{12x}\right)
\\&=\color{red}{0}
\end{aligned}$$
HINT:  $$\left(\sqrt[\beta]{1+\frac{1}{x}}-1\right)\approx_\infty \frac{1}{x} \times \frac{1}{\beta}$$
