How to solve asymptotic expansion: $\sqrt{1-2x+x^2+o(x^3)}$

Question

Determinate the best asymptotic expansion for $x \to 0$ for:

$$\sqrt{1-2x+x^2+o(x^3)}$$

How should I procede?
In other exercise I never had the $o(x^3)$ in the equation but was the maximum order to consider.

Answer 1

Hint. One may recall the Taylor series expansion, as $u \to 0$, $$ \sqrt{1+u}=1+\frac{u}2-\frac{u^2}8+\frac{u^3}{16}+o(u^3). $$ Setting, as $x \to 0$, $$ u=-2x+x^2+o(x^3) $$ it gives $$ 1+\frac{-2x+x^2+o(x^3)}2-\frac{(-2x+x^2+o(x^3))^2}8+\frac{(-2x)^3}{16}+o(x^3) $$ or $$ 1+\frac{-2x+x^2}2-\frac{4x^2-4x^3}8+\frac{-8x^3}{16}+o(x^3) $$thus, as $x \to 0$,

$$ \sqrt{1-2x+x^2+o(x^3)}=1-x+o(x^3). $$

Answer 2

You have the following asymptotic expansion :

$$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}+o(x^3)$$

So :

$$\sqrt{1-2x+x^2+o(x^3)}=1+\frac{-2x+x^2+o(x^3)}{2}-\frac{(-2x+x^2+o(x^3))^2}{8}+\frac{(-2x+x^2+o(x^3))^3}{16}+o((-2x+x^2+o(x^3))^3)\\=1+\frac{-2x+x^2}{2}-\frac{4x^2-4x^3+x^4}{8}+\frac{-8x^3+12x^4-6x^5+x^8}{16}+o(x^3)\\=1-x+\frac{x^2}{2}-\frac{x^2}{2}+\frac{x^3}{2}-\frac{x^3}{2}+o(x^3)=1-x+o(x^3)$$