Intuition on what $(1-o(1))$ means.

Question

Does $(1-o(1))$ imply something very close to $1$ or something very close to $0$. I understand that $o(1)$ means a function growing at less than 1. Thanks for your help.

Answer 1

A function $f\in 1-o(1)$ iff a function $g\in o(1)$ exists satisfying $f(x)\equiv1-g(x)$. So in the limit where $g\to0$, $f\to1$.

Answer 2

$f(x) = o(g(x))$ is defined by saying that for any positive $\epsilon$, there exists a value $N$ such that $\forall x \geq N \ |f(x)| \leq \epsilon g(x)$, i.e. $f$ is bounded by $\epsilon g$.

So if $1 - f(x) = o(g(x))$ where $g(x) = 1$, we have that $\forall \epsilon > 0$ there is a value $N$ such that $\forall x \geq N \ |1 - f(x)| \leq \epsilon$, which implies that $\lim_{x \rightarrow \infty} f(x) = 1$.

Answer 3

Instead of Landau notation $ o(g(x)) $ near the point $ x=a $, it is more pratical to write $$g(x)\epsilon(x)\;with\; \lim_{x\to a}\epsilon(x)=0$$

So

$$1-o(\color{red}{1})=1-\color{red}{1}.\epsilon(x)=1-\epsilon(x)$$ and

$$2-o(x^3)=2-x^3\epsilon(x)$$

Answer 4

Let me add one more view: $o(1)$ is set of functions with limit zero i.e. $f\in o(1)\Leftrightarrow \lim f = 0$. Also set is expression $1 \pm o(1)=\{g\colon g=1\pm f, f \in o(1) \}$.

So we can conclude, that $g=1 \pm o(1)$ is equivalence $\lim g =1$.