if I have a function $f$ of the form
$f(x) = \frac{1}{1 + O(x^k)}$ for a given real number $k$
then can I say, by expansion of Taylor series, that
$f(x) = \sum_{n=0}^\infty \binom{-1}{n}O(x^k)^n$
and since f decays at least as fast as the first order term of the binomial expansion, I can say that $f(x) \approx 1 + O(x^k)$ in the limit process of $x$ that tends to $0$ ??
Thanks to ronno and GEdgar for the tips in the comment section, I tried to do the same by means of Big-O definition, as $x \to 0$, so here's my attempt
We want to show that if $f(x) = \frac{1}{1+O(x^k)}$ then $f(x) = 1 + O(x^k)$ as $x \to 0$
Let's define $h(x) = f(x)-1$, so now we have to show that $h(x) = O(x^k)$ or equivalently, employing the definition of the big-O notation, that
$\limsup_{x\to 0}\frac{|h(x)|}{x^k} < \infty $
where $h(x) = -\frac{O(x^k)}{1+O(x^k)}$
but
$\limsup_{x\to 0}\frac{|h(x)|}{x^k} = \limsup_{x\to 0}\frac{1}{1+O(x^k)} = 1 $
Hence $f(x) = 1 + O(x^k)$.
Is this correct ?
Perhaps it would be clearer to use the following definition of the Big-O:
$$g(x)=O(h(x)) \quad \text{as} \quad x\to 0$$
if and only if there exists $M>0$ and $\epsilon>0$ such that $|g(x)|\leq M|h(x)|$ for all $x$ satisfying $|x|<\epsilon$.
In your case you have $f(x) = \frac{1}{1 + g(x)}$ with $g(x)=O(x^k)$. Let $M,\epsilon>0$ be such that $|g(x)|\leq M|x^k|$ whenever $|x|<\epsilon$. By choosing $\epsilon>0$ smaller if necessary, you can ensure that $|g(x)|<\delta$ whenever $|x|<\epsilon$ for some $\delta<1$. Then $$|f(x)-1|=\frac{|g(x)|}{|1+g(x)|}\leq \frac{M|x^k|}{1-\delta}$$ whenever $|x|<\epsilon$. Hence $f(x)-1=O(x^k)$, i.e. $f(x)=1+O(x^k)$.