Mathematica can't simplify asymptotic expressions containing constant symbols

Question

I want to calculate simple asymptotic expressions involving positive constant symbols ($a > 0$), such as $$\lim_{x\to\infty} \operatorname{sech}(a x) \sim 2 e^{-a x}$$

Surprisingly, the Asymptotic function of Mathematica can't calculate this limit. The code

Assuming[a > 0, Asymptotic[Sech[a x], x -> ∞]]

returns Sech[a x]

while

Asymptotic[Sech[3 x], x -> ∞]

correctly returns 2 E^(-3 x)

How can I get Mathematica to evaluate this asymptotic limit correctly?

Edit 1:

One hack is to replace $a$ with $\pi$, then calculate the asymptotic limit, then convert $\pi$ back to $a$.

Asymptotic[Sech[a x] /. a -> π, x -> ∞] /. π -> a

returns the desired limit 2 E^(-a x)

Edit 2:

An actual example I am working on is

Assuming[a > 0, Asymptotic[-(1/2) Sech[1/2 x Sinh[a]]^2 Sinh[a]^2, x -> Infinity]]

So far only the $\pi$ trick works for this case, giving -2 E^(-x Sinh[a]) Sinh[a]^2.

Answer 1

TrigToExp does the job

Assuming[a > 0, Asymptotic[TrigToExp[Sech[a *x]], x -> Infinity]]

2 E^(-a x)

Assuming[a \[Element] Reals, Asymptotic[TrigToExp[Sech[x/a] + Sech[x*a]], x ->Infinity]]

ConditionalExpression[2 E^(-(x/a)), a > 1 && a^2 < 3]

Answer 2

Try this:

Simplify[Refine[Asymptotic[Sech[a x], a x -> ∞], 
Assumptions -> Element[a,Reals]]]

Answer 3

It appears that Mathematica simply cannot handle the asymptotics of Sech in general. When requesting the first order term, if you give it TrigToExp@Sech[a x] $= \dfrac{2}{e^{-a x}+e^{a x}}$ it's smart enough to deal with those exponentials. But if you try

Asymptotic[TrigToExp@Sech[a x], x -> ∞, Assumptions -> a > 0, SeriesTermGoal -> 2]

you'll find this doesn't work. This suggests that Mathematica can't actually generate the asymptotic expansion needed to deal with this. So for such things you'll just have to try to massage them into a form that Mathematica can digest.

For example, the next problem you left in a comment, $$ -\frac{1}{2} \sinh ^2(a) \text{sech}^2\left(\frac{1}{2} x \sinh (a)\right) $$ can be handled with

FullSimplify@Assuming[a>0,
Asymptotic[ExpandAll@TrigToExp[-(1/2) Sech[1/2 x Sinh[a]]^2 Sinh[a]^2],x->∞]]

which gives back $$ 2 \sinh ^2(a) (\sinh (x \sinh (a))-\cosh (x \sinh (a))) $$