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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 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".2011-07-28

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Apparently, there is no standard terminology for that, but:

Quoting from this paper (p. 4):

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.

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    I hadn't seen that paper before. Thanks. I agree that terminology isn't great.2011-07-28