What is $(1 + O(h))^{-1}$?

Question

I need some help figuring this one out:

Is $(1 + O(h))^{-1}$ also equal to $1 + O(h)$?

I know that $$\lim_{h \to 0} \frac{O(h)}{h} = C$$ for some $C \in \Bbb R$.

I can show that $$ \frac{1}{1 + O(h)} - 1 = O(h).$$

We have $$ \lim_{h \to 0} \frac{1}{h}\left(\frac{1}{1 + O(h)} - 1\right) = \lim_{h\to 0} \frac{1}{h}\frac{-O(h)}{1 + O(h)} = \lim_{h\to 0}-\frac{C}{1+O(h)}.$$

Answer 1

First things first, $f(h)=O(h)$ when $h\to 0$ means $|f(h)|\le C|h|$ when $h$ is in some neighbourhood of $0$. The condition you are citing ($\lim_{h\to 0}\frac{f(h)}{h}=C$) is just a sufficient condition for that to happen.

Still, the statement you want to prove is true: if $|f(h)|\le C|h|$ in some neighbourhood of $0$, then:

$$|(1+f(h))^{-1}-1|=|f(h)||(1+f(h))^{-1}|\le C(1+\epsilon)|h|$$

in some (potentially smaller, depending on your choice of $\epsilon\gt 0$) neighbourhood of $0$: the factor $1+\epsilon$ comes from approximating $(1+f(h)^{-1})\to 1$ when $h\to 0$.

Therefore, $(1+f(h))^{-1}-1=O(h)$ (take $C(1+\epsilon)$ as the constant multiplier), i.e.$(1+f(h))^{-1}=1+O(h)$.

Answer 2

It is true that $(1+O(h))^{-1}=1+O(h)$ as $h\to 0$. There is a remark to be done, and it is that the notation $f(h)=O(h)$ usually means $|f(h)|\le C|h|$ for all $h$ in a neighborhood of $0$, or equivalently, $$ \limsup_{h\to 0} \frac{|f(h)|}{|h|}=C\in\mathbb R, $$ as G. Sassatelli points out.

With this definition you cannot use your argument. But there is an even easier one based on the Taylor expansion of the function $f(x):=(1+x)^\alpha$, where $\alpha \in \mathbb C$. That is,
$$ (1+O(h))^\alpha = 1+\alpha O(h)+ O(h^2)=1+O(h).$$ Notice that this works for all values of $\alpha$, even complex ones.