What is little oh and big oh formally?

Question

I'm reading a text and it says the following for a function $c: \mathbb{R}^d \to \mathbb{R}$:

Suppose that $c(x) = 1 + o(\Vert x \Vert)$ if $x \to 0$ and $c(x) = O(1/\Vert x \Vert)$ if $\Vert x \Vert \to \infty$. Can someone explain what this means formally? I could not find a definition in the book.

Google seems to suggest that

$$c(x) = O(1/\Vert x\Vert), \Vert x \Vert \to \infty$$ means that there is $M \geq 0$, $a \geq 0$ such that $\Vert x \Vert \geq a \implies |c(x)| \leq M/\Vert x \Vert$. Is that correct?

And what about $c(x) = 1+ o(\Vert x \Vert ), x \to 0$. What does it formally mean?

Answer 1

Suppose that $f:\mathbb R^n \to \mathbb R$ and $g:\mathbb R^n \to \mathbb R$. Let $a \in \mathbb R^n$. To say that $f$ is $o(g)$ as $x$ approaches $a$ means that if $C > 0$ then $|f(x)| \leq C |g(x)|$ for all $x$ sufficiently close to $a$. To be more precise, it means that if $C > 0$ then there exists $\delta > 0$ such that if $\|x - a \| < \delta$ then $|f(x)| \leq C |g(x)|$.

In the special case that $g(x) = \| x \|$, the statement that $f$ is $o(g)$ as $x$ approaches $0$ means that if $C > 0$ then there exists $\delta > 0$ such that if $\|x \| < \delta$ then $|f(x)| \leq C \|x \|$. But this is equivalent to saying that $$ \frac{f(x)}{\|x\|} \to 0 \quad \text{as } x \to 0. $$

Finally, to say that $c(x) = 1 + o(\|x\|)$ as $x$ approaches $0$ means that the function $c - 1$ is $o(\|x\|)$ as $x$ approaches $0$. In other words, $$ \frac{c(x) - 1}{\| x\|} \to 0 \quad \text{as } x \to 0. $$

Answer 2

1. Google is correct.

2. We have

$$c(x) = 1+ o(\Vert x \Vert ), x \to 0$$

$ \iff$

$$\frac{c(x)-1}{\Vert x \Vert} \to 0, x \to 0 .$$

Answer 3

The definition is $f\in o(g)$ as $x\to a$ if and only if for all $c>0$ there is some $\delta>0$ such that $\lvert f(x)\rvert<c\lvert g(x)\rvert$ for all $x$ such that $\lVert x-a\rVert<\delta$.

As for $O$, the definition is $f\in O(g)$ as $x\to a$ if and only if there are some $c>0$ and $\delta>0$ such that $\lvert f(x)\rvert<c\lvert g(x)\rvert$ for all $x$ such that $\lVert x-a\rVert<\delta$.