Can't solve following limit: $\lim_{x \to \infty}x \left( \sqrt[3]{5+8x^3} - 2x\right)$

Question

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.

Answer 1

Hint: $$\sqrt[3]{5+8x^3} - 2x = \sqrt[3]{5+8x^3} - \sqrt[3]{8x^3} = \frac{\left(5+8x^3\right)-8x^3}{\left(\sqrt[3]{5+8x^3}\right)^2 + \sqrt[3]{5+8x^3} \cdot \sqrt[3]{8x^3} + \left(\sqrt[3]{8x^3}\right)^2}.$$

Answer 2

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.

Answer 3

Putting $n=\frac1h$

$$\lim_{x \to \infty}x \left( \sqrt[3]{5+8x^3} - 2x\right) $$

$$=\lim_{h\to0}\frac{\sqrt[3]{5h^3+8}-2}{h^2}$$

$$=\lim_{h\to0}\frac{5h^3+8-2^3}{h^2\{(5h^3+8)^{\frac23}+(5h^3+8)^{\frac13}+1\}}$$

$$=\lim_{h\to0}\frac{5h}{\{(5h^3+8)^{\frac23}+(5h^3+8)^{\frac13}+1\}}$$

$$=\cdots$$