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
 A: I'm not sure if it's elegant in your eyes but it's at least efficace
$$(x^3+2x)^\frac{1}{3}-\sqrt{x^2-2x}=x\left(1+\frac{2}{x^2}\right)^{1/3}-x\left(1-\frac{2}{x}\right)^{1/2}\sim_\infty 1$$ 
using that
$$(1+x)^\alpha\sim_0 1+\alpha x$$
A: 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:
$$\sqrt[3]{x^3+2x}-\sqrt{x^2-2x}=(\sqrt[3]{x^3+2x}-x)-(\sqrt{x^2-2x}-x)\to 1$$.
A: It can be done by conjugation, but it is not pleasant to type.  We sketch the approach. We are looking at
$$((x^3+2x)^2)^{1/6} -((x^2-2x)^3)^{1/6}=u-v.$$
Multiply top and missing bottom by $u^5+u^4v+u^3v^2+u^2v^3+uv^4+v^5$. 
On top we get $u^6-v^6$, a polynomial with lead term $6x^5$. At the bottom, we get $6$ terms each of which asymptotically behaves like $x^5$. Dividing top and bottom by $x^5$ and manipulating a bit gives us limit $1$. 
Remark: Looking at the beginning of the Taylor expansions is the "right" way to do the problem.
