How to solve this type of exercises $\sqrt{x^6+x^5-2x^3+O(x^2)}$

Question

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?

Answer 1

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 $$\sqrt{x^6+x^5-2x^3+O(x^2)}=\sqrt{x^6+x^5-2x^3}+O(x^{-1})$$