Why $\lim_{x \to 0} \frac{O(x)}{x}=0$?

Question

Why does $\lim_{x \to 0} \frac{O(x)}{x}=0$? Here $O$ means big $O$ notation.

Answer 1

This is untrue. The limit, to the contrary can be as large as you want. For example, let $C \in \mathbb{R}_+$ be "large." Then $f(x)=Cx$ is $O(x)$. Yet,

$$\lim_{x \to 0} \frac{Cx}{x}=C$$

More formally: $f$ is $O(x)$ as $x \to \infty$ $\iff$ there exist $M \in \mathbb R$ and $x_0$ so that $x \geq x_0$ where $f(x) \leq M|x|$ for all $x \geq x_0$.

Sometimes, $O(x)$ is meant to approximate $x$ locally, say at $0 \in \mathbb{R}$. Here, we can reformulate the definition to mean the following: $f$ is $O(x)$ at $x_0$ $\iff$ there exist $M \in \mathbb R$ and $\epsilon$ so that $x \in B(x_0,\epsilon)$ implies that $f(x) \leq M|x|$.

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Here is a true statement:

if $f(x)$ is $O(x)$ at $0$, then $\lim_{x \to 0} f(x)=0$. Why is this true? Well, by assumption there exists some $M \in \mathbb R$ so that $|x|<\delta$ implies that $f(x) \leq M|x|$. But, we know that $x \to 0$, meaning that $$-M \cdot \lim_{x \to 0} \,x \leq\lim_{x \to 0}f(x) \leq M \cdot \lim_{x \to 0}\, x \implies \lim_{x \to 0} f(x)=0$$