In perturbation theory, why is nothing asymptotic to zero?

Question

I'm a physicist who's recently begun studying asymptotics and perturbation theory, and I'm confused as to why nothing can be asymptotic to zero (in the sense of $f(x)\sim 0$ as $x\to x_0$).

Answer 1

In short: because then, $f$ has to be identically $0$ on some open interval around $x_0$, and that's not a very interesting situation...

To see why: Let's have a look at the definition of asymptotics:

We have that $$ f(x) \sim_{x\to x_0} g(x) $$ means $$ f(x)-g(x) =_{x\to x_0} o(g(x)) $$

which in turn means

There exists a non-negative function $\varepsilon$ (defined on a neighborhood $U$ of $x_0$) such that $\lim_{x\to x_0} \varepsilon(x) = 0$ and $\lvert f(x) - g(x)\rvert \leq \varepsilon(x) \lvert g(x)\rvert$ for all $x\in U$.

Now, that implies that if $f(x) \sim_{x\to x_0} 0$, one must have $\varepsilon$ and $U$ as above such that $$ \lvert f(x) \rvert\leq \varepsilon(x)\cdot 0 = 0, \qquad \forall x\in U $$ and consequently $f$ has to be identically $0$ on $U$.