# How can I investigate limit at infinity with asymptotic series?

I have the following function $$h_1(x)$$:

h1[x : _] = (1/(8 a))Sech[a x]^2 (-12 a^3 + 8 a d + 8 a c x + 96 a^3 Log[Cosh[a x]] +
12 a^3 Cosh[3 a x] Sech[a x] +
c Sech[a x] Sinh[
3 a x] - (c + 4 a^2 (4 b + x (12 a^2 + 2 d - 12 a^3 x + c x)) +
96 a^4 x (-Log[1 + E^(-2 a x)] + Log[Cosh[a x]]) +
48 a^3 PolyLog[2, -E^(-2 a x)]) Tanh[a x])


I want to choose $$b$$, $$c$$ and $$d$$ to make $$h_1$$ go to zero at infinity.

Is there a way to do this with asymptotic series in Mathematica?

• @user64494, please thoroughly and carefully read instructions for editing posts. Especially pay attention to: Tiny, trivial edits are discouraged. Jan 8 at 18:24
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Here a solution using Asymptotic

h[x_] := (1/(8 a)) Sech[a x]^2 (-12 a^3 + 8 a d + 8 a c x +
96 a^3 Log[Cosh[a x]] + 12 a^3 Cosh[3 a x] Sech[a x] +
c Sech[a x] Sinh[
3 a x] - (c + 4 a^2 (4 b + x (12 a^2 + 2 d - 12 a^3 x + c x)) +
96 a^4 x (-Log[1 + E^(-2 a x)] + Log[Cosh[a x]]) +
48 a^3 PolyLog[2, -E^(-2 a x)]) Tanh[a x])


find asymptote

asy = Asymptotic[h[x], x -> Infinity] (*ConditionalExpression[6 a^2 + c/(2 a), (b | c | d) \[Element] Reals && a > 0]*)


should be zero

solc = Solve[asy[[1]] == 0, c][[1]] (* {c -> -12 a^3} *)


solution h[x]

h[x]/.solc//Simplify (* depends on parameter a>0, b,d)


higher order approximations follow step by step

asy2 = Asymptotic[h[x] /. solc, x -> Infinity ] // Simplify
(* ConditionalExpression[4 a E^(-2 a x)x (-d + 6 a^2 (-1 + Log[4])),(b | d) \[Element] Reals && a > 0] *)

h[x] /. solc /. d -> 6 a^2 (-1 + Log[4]) // Simplify


• +1. I found the same condition by hand with help of Mathematica. The first step was Expand[(TrigToExp[(1/(8 a)) Sech[a x]^2 (-12 a^3 + 8 a d + 8 a c x + 96 a^3 Log[Cosh[a x]] + 12 a^3 Cosh[3 a x] Sech[a x] + c Sech[a x] Sinh[ 3 a x] - (c + 4 a^2 (4 b + x (12 a^2 + 2 d - 12 a^3 x + c x)) + 96 a^4 x (-Log[1 + E^(-2 a x)] + Log[Cosh[a x]]) + 48 a^3 PolyLog[2, -E^(-2 a x)]) Tanh[a x])] /. {Exp[-a*x] -> 0, Exp[-2*a*x] -> 0, Exp[-3*a*x] -> 0}) /. Log[E^(a x)/2] -> a*x - Log[2]]. Jan 8 at 18:11
• @simon Hope it helps! You are welcome! Jan 8 at 18:21
• Expand[(TrigToExp[h1[x]] /. {Exc = -12 a^3p[-ax] -> 0, Exp[-2*ax] -> 0, Exp[-3*ax] -> 0}) /. Log[E^(a x)/2] -> ax - Log[2]] Thank you so c = -12 a^3 but the following still draws an echo not a limit of 0 Evaluate[Limit[h1[x], x -> Infinity]] Jan 8 at 18:57
• @simon See the comments of the first two commands. I'm unsure whether Asymptotic is known in v9 Jan 8 at 19:27
• @simon: Did you pay you attention to the words "The first step was ..." in my comment? Jan 8 at 19:39