# Asymptotics where the absolute error goes to 0 - what are these called?

Say I have a function $f$ for which $\lim_{x\rightarrow\infty}f(x)=\infty$ and which I'd like to approximate by a simpler function $g$. We say $g$ is an asymptotic for $f$ iff $$\lim_{x\rightarrow\infty} \frac{f(x)}{g(x)} = 1,$$ i.e., iff the values $f(x)$ can be replaced with the values $g(x)$ with a percentage error that goes to $0$ as $x\rightarrow\infty$.

Sometimes the asymptotic $g(x)$ is even better, in the sense that the absolute error (not just the percentage error) goes to $0$ as $x\rightarrow\infty$, i.e., we have $$\lim_{x\rightarrow\infty}(f(x)-g(x)) = 0,$$ or said another way, $$f(x) = g(x) + o(1).$$

My question is this: Do such asymptotics have a special name? To give a silly example, consider $f(x)=x^2+1/x$. We have that $g(x)=x^2+x$ is an asymptotic for $f$, but it doesn't meet the second condition. A better asymptotic (one that does meet the second condition) would just be $g(x)=x^2$.

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I doubt it; this is an extremely rare condition and I don't think it is likely to show up in practice. – Qiaochu Yuan Jul 28 '11 at 3:41
The connection to numerical-methods is tenuous at best, so I've taken it down. If anybody thinks it's relevant here, feel free to add it back. – J. M. Jul 28 '11 at 3:42
@Qiaochu Yuan: I'm writing a paper right now where I develop asymptotics that have this property. At first I was just trying to find asymptotics, but then I noticed the asymptotics I was getting had the second property as well. One example where this kind of thing comes up in practice is the average number of components in a supercritical sequence construction (see p. 294 of the book Analytic Combinatorics, freely available online), but they don't use any special terminology. – Martin Malandro Jul 28 '11 at 3:49
I don't know a name for it, but it says $e^f$ and $e^g$ are asymptotic, so I suppose you could say the functions are "exponential-asymptotic". – Gerry Myerson Jul 28 '11 at 4:15

Asymptotic notation. We abbreviate $\lim _{t \to \infty } [f(t) - g(t)] = 0$ by $f(t) \stackrel{t \to \infty}{\longrightarrow} g(t)$ and say $f$ converges to $g$, without implying that $\lim _{t \to \infty } g(t)$ itself exists.