Solve $\sqrt{1-2x+x^2+o(x^3)}$ with $x \to 0$

Question

I need help to solve $\sqrt{1-2x+x^2+o(x^3)}$ with $x \to 0$, I do not understand when and why I should stop.

Here my steps:

I can use Taylor formula for $\sqrt{1+t}$ so:

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

where: $t = -2x+x^2+o(x^3)$

I start with the first order:

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

This do not solve the problem so I think I should continue wih the second order.
Using the same steps above I get:

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

is it correct?
At this point I don't know how to procede, I don't think that this solve the problem, should I continue to the 3rd order?

Answer 1

You are dealing with this limit $$\lim_{x \rightarrow 0} \sqrt{1-o(x)}.$$ As $x$ tends to become zero, all terms containing an $x$ will also tend to zero. However, the initial term with no $x$ term associated will simply remain one. Hence, the value of the limit will be simply $\sqrt{1}$.

Answer 2

We have $\sqrt{1-2x+x^2} = 1 - x$. Now $\sqrt{}$ is a Lipschitz function in a neighbourhood of $1$, so $\sqrt{1-2x+x^2 + o(x^3)} = \sqrt{1-2x+x^2} + o(x^3) = 1 - x + o(x^3)$.