Taylor series $\sqrt{1-2x+x^2+o(x^3)}$ I need to solve this asymptotic expansion for $x\to0$
$$\sqrt{1-2x+x^2+o(x^3)}$$
This is the expansion to use:
$$\sqrt{1+t} = 1+ \frac{t}{2}-\frac{t^2}{8}+\frac{t^3}{16}+o(x^3)$$
let:
$$t = -2x+x^2+o(x^3)$$
My question is, should I stop to the first order or keep going?
How can I decide when to stop?
I have tried first and second order and I get:
$$\sqrt{1+t}= 1+ \frac{-2x+x^2+o(x^3)}{2}+o(-2x+x^2+o(x^3) = 1-x+\frac{1}{2}x^3+o(x^3)$$
The second grade:
$$\sqrt{1+t}= 1+ \frac{-2x+x^2+o(x^3)}{2} - \frac{\big(-2x+x^2+o(x^3)\big)^2}{8} +o\big(-2x+x^2+o(x^3)\big)^2 = 1-x+4x^2+\frac{1}{2}x^3-4x^3-\frac{1}{8}x^4+x^4+o(x^4)$$

Since we have an $o(x^3)$ term we can use the following
$$\sqrt{1+t} = 1+ \frac{t}{2}-\frac{t^2}{8}+\frac{t^3}{16}+o(t^3)$$
that is
$$\sqrt{1-2x+x^2+o(x^3) }=1+ \frac{-2x+x^2+o(x^3)}{2}-\frac{(-2x+x^2+o(x^3))^2}{8}+\frac{(-2x+x^2+o(x^3))^3}{16}+o((-2x+x^2+o(x^3))^3)$$
keeping only the term with order less than $o(x^3)$, that is
$$1-x+\frac12x^2 -\frac12x^2+\frac12x^3 -\frac12x^3+o(x^3) =1-x+o(x^3)$$
