How to solve this type of exercises $\sqrt{x^6+x^5-2x^3+O(x^2)}$ I have a simulation test with this type of exercise, asymptotic expansion:
$$\sqrt{x^6+x^5-2x^3+O(x^2)}$$
with
$$ x\rightarrow \infty$$
I have studied the theory of Landau's symbols but I have no idea about how to solve.
Can someone please explain me how to do that? 

Write
$$\sqrt{x^6+x^5-2x^3+O(x^2)}=$$
$$=\sqrt{x^6+x^5-2x^3}+\sqrt{x^6+x^5-2x^3+O(x^2)}-\sqrt{x^6+x^5-2x^3}=$$
$$=\sqrt{x^6+x^5-2x^3}+\frac{O(x^2)}{\sqrt{x^6+x^5-2x^3+O(x^2)}+\sqrt{x^6+x^5-2x^3}}.$$
But $\sqrt{x^6+x^5-2x^3+O(x^2)}+\sqrt{x^6+x^5-2x^3}\sim 2x^3$, hence
$$\frac{O(x^2)}{\sqrt{x^6+x^5-2x^3+O(x^2)}+\sqrt{x^6+x^5-2x^3}}=O(x^{-1})$$
thus
$$\sqrt{x^6+x^5-2x^3+O(x^2)}=\sqrt{x^6+x^5-2x^3}+O(x^{-1})$$
