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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!

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    Order terms assume a certain limit value for the variable (x), which may not necessarily be 0. It could be that in this context, the limit refers to $x \to \infty$ instead. – 2017-01-29
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    (After edit) I don't understand your last paragraph. If the context is indeed that we are considering the limit $x \to \infty$, then everything is fine, no? You can find a $K > 0$ such that the ratio is bounded by $K$ as $x \to \infty$. – 2017-01-29
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    @TMM I'm looking for a reference to the second definition. preferably in a book – 2017-01-29
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    "'m looking for a reference to the second definition. preferably in a book" And WP is not an option, because? – 2017-01-29
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    I agree with Did, this is such a standard concept that there's no reason to look for a book with this definition. Just accept the fact that you incorrectly assumed the limit had to be for $x$ approaching $0$ and move on. – 2017-01-29

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$f(x)=O(g(x))$ 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 $-\infty.$