= Asymptotic equivalence
{title2=$a_n\sim b_n\iff a_n/b_n\to1$}
= Asymptotically equivalent
{synonym}
For <functions> with $g$ nonzero sufficiently near a specified <limit>,
$$
\boxed{f\sim g\quad\Longleftrightarrow\quad \frac{f}{g}\longrightarrow1.}
$$
The same definition applies to <sequences> $a_n\sim b_n$ as $n\to\infty$. Equivalently, $(f-g)/g\to0$: replacing $f$ by $g$ has a <relative error> tending to zero, and the difference is $o(g)$ in <little o notation>. The <limit> and reference scale are essential parts of the statement.
For example, $n^2+n\sim n^2$ because $(n^2+n)/n^2=1+1/n\to1$. Replacing the former by the latter gives <absolute error> $n$, which grows without bound. Thus <asymptotic equivalence> controls a ratio, rather than requiring the <absolute error> to vanish. This distinction is crucial when approximating a rapidly growing <factorial> with <Stirling's formula>.
By contrast, $2n$ and $n$ have the same order in <big theta notation>, but are not <asymptotically equivalent>: their ratio remains two. For positive <functions>, <asymptotic equivalence> gives both-sided constant bounds eventually, whereas constant bounds alone do not recover the leading coefficient.
\Image[/asymptotic-local-exponential-errors.png]
{title=Local exponential approximation: relative difference tends to zero and absolute error is quadratic}
For a local example with real $h$, the <exponential function> obeys $e^h-1\sim h$ as $h\to0$. The <Taylor theorem with Lagrange remainder> gives
$$
e^h-1=h+\frac{e^\xi}{2}h^2,\qquad
|\xi|\le |h|.
$$
Thus for $|h|\le1$,
$$
|e^h-1-h|\le\frac e2 h^2.
$$
This states $e^h-1=h+O(h^2)$ in <big O notation>, and division by $h$ shows that the relative difference measured against $h$ vanishes. The diagram uses positive $h$; the bound is valid on both sides of zero. At $h=0.1$, the <absolute error> is approximately $0.00517092$ and the relative difference against $h$ is approximately $0.0517092$.
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