computing $\lim_{X \to \infty }((x^3+2x)^\frac{1}{3}-\sqrt{x^2-2x})$ $\lim_{X \to \infty }((x^3+2x)^\frac{1}{3}-\sqrt{x^2-2x})$ 
Anyone have a nice elegent way wo solve it? i tried L'Hôpital's rule but didn't work.
also i tried to multiple by Conjugation and it failed too.
thanks

We have:
$$\sqrt[3]{x^3+2x}-x= x\Bigl(\sqrt[3]{1+\frac 2{x^2}}-1\Bigr)\sim \frac 2{3x}\to 0$$
$$\sqrt{x^2-2x}-x= x\Bigl(\sqrt{1-\frac 2x}-1\Bigr)\to -1$$
Consequently:
$$\sqrt[3]{x^3+2x}-\sqrt{x^2-2x}=(\sqrt[3]{x^3+2x}-x)-(\sqrt{x^2-2x}-x)\to 1$$.
