Asymptotic Notations: $\sim$ vs $\asymp$ and PNT Is the only difference between $f(x) \sim g(x)$ and $f(x)\asymp g(x)$ the constant which their ratio approaches?
My understanding is that $f(x)\asymp g(x)$ means that $f(x)=O(g(x))$ and $g(x)=O(f(x))$, and that this is equvialent to saying that
$$\lim\frac{f(x)}{g(x)}$$
exists (or at least, $\limsup$) and is some fixed number.
On the other hand, $f(x)\sim g(x)$ means that this limit is specifically $1$.
I ask this question because it doesn't feel like one is that much stronger than the other, they both establish that $f(x)$ and $g(x)$ are the same "kind" of function with respect to their growth (they feel more or less the same qualitatively), $\asymp$ is more akin to "proportional to" in the limit, rather than $\sim$ which is sort of like "equal to" in the limit.
In particular, I'm surprised how easy the proof of Chebychev's theorem $\pi(x)\asymp x/\log x$ is when compared with the PNT $\pi(x)\sim x/\log x$.
 A: There is a very large difference between $f(x) \asymp g(x)$ (in the traditional meaning that there are constants $0 < c < C < \infty$ such that $c g(x) \leq f(x) \leq C g(x)$, sometimes stated with the additional caveat that this only needs to hold for all sufficiently large $x$) and $f(x) \sim g(x)$ (which typically means that $\lim f(x) / g(x) = 1$.
For example, consider $f(x) := 1 + 0.5 \chi_\mathbb{Q}(x)$, where
$$
\chi_\mathbb{Q}(x) = \begin{cases} 1 & x \in \mathbb{Q}, \\ 0 & \text{else} .\end{cases}
$$
This is a terribly discontinuous function, but nonetheless $f(x) \asymp 1$. For that matter, $f(x) \asymp 100$ too.
A: Call $h=\frac{f}{g}$. Then $f \asymp g$ means $h$ is bounded, while $f\sim g$ means $h$ has a limit (and that limit is $1$).
Those are VERY different things.
That's like comparing testing the convergence of a series with finding its sum.
Proving the convergence of $$\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^42^k{2k \choose k}}$$
is very easy, finding that its sum is
$$4\operatorname{Li}_4\left(\frac12\right)-\frac72\zeta(4)+\frac{13}4\ln2\zeta(3)-\ln^22\zeta(2)+\frac5{24}\ln^42$$
is not.
