I recently saw in a paper on optimization, it was written that (given $f$ as function typically found in single variable calculus)
$f(x) + c_1x+c_2 = f(x) + \mathcal{O}(x)$
In optimization, the definition of $\mathcal{O}(x)$ as in Luenberger is given as:
If $h$ is a real valued function of a real variable, the notation $h(x) = \mathcal{O}(x)$ means that there exists a $K \geq 0$ such that $\Big|\dfrac{h(x)}{x}\Big| \leq K$ as $x \to 0$
But clearly here $\Big|\dfrac{c_1x+c_2}{x}\Big| \to \infty$ as $x \to 0$? Wouldn't it be the case such $K$ would not exist?
But if I were to interpret $\mathcal{O}(x)$ directly, it means that there exists some function $K_1x + K_2$ such that $K_1x + K_2 \geq c_1x+c_2$ as $x \to \infty$. Clearly I need a better definition!
$f(x)=O(g(x))$ as $x\to \infty$ means there exists $k$ such that $|f(x)|\leq k|g(x)|$ for all sufficiently large $x.$
$f(x)=O(g(x))$ as $x\to 0$ means there exists $k$ such that $|f(x)|\leq k|g(x)|$ for all sufficiently small $|x|.$
Unless it is obvious from the context, it must be specified where $x$ is going, just as $\lim h(x)=A$ is unclear unless it is specified where $x$ is going.
In the paper it must mean that $x$ is going to $\infty$ or to $-\infty.$