Grouping expressions by their $O(..)$ rate of growth around 0

Question

Given a list of expressions like $3 x^2,\frac{\sqrt{x}}{\sqrt{2}},\frac{2 \sqrt{x}}{\pi },\frac{9 x^3}{2},x,$, how could I group them based on their rates of growth around 0?

IE, $2x$ and $3x$ are both $O(x)$, while $x^2$ is $O(x^2)$.

Code below generates 25 expressions I'm trying to group in this way:

pairs = {
   {"Kumaraswamy", KumaraswamyDistribution[2, 3]},
   {"Weihbul", WeibullDistribution[1/2, 2]},
   {"ArcSin", ArcSinDistribution[]},
   {"Bates", BatesDistribution[3]},
   {"Uniform", UniformDistribution[]},
   {"Pareto", ParetoDistribution[1, 2]},
   {"Logistic", LogisticDistribution[]},
   {"Extreme Value", ExtremeValueDistribution[]},
   {"Frechet", FrechetDistribution[2, 1, 0]},
   {"Erlang", ErlangDistribution[3, 2]},
   {"Gamma", GammaDistribution[2, 3]},
   {"InverseGamma", InverseGammaDistribution[2, 3]},
   {"Gumbel", GumbelDistribution[0, 1]},
   {"Beta", BetaDistribution[1/2, 1]},
   {"Marchenko-Pastur", MarchenkoPasturDistribution[1]},
   {"Semicircle", WignerSemicircleDistribution[1]},
   {"LogNormal", LogNormalDistribution[0, 1]},
   {"Cauchy", CauchyDistribution[]},
   {"Pareto", ParetoDistribution[1, 2]},
   {"Normal", NormalDistribution[]},
   {"Student-T", StudentTDistribution[1]},
   {"ChiSquared", ChiSquareDistribution[1]},
   {"Chi", ChiDistribution[1]},
   {"Exponential", ExponentialDistribution[1]},
   {"Inverse Normal", InverseGaussianDistribution[2, 1]}
   };
trunc[dist_] := TruncatedDistribution[{0, \[Infinity]}, dist];

getGrowth[dist_] := 
  Assuming[0 < x < 1/1000, Asymptotic[Refine@CDF[dist, x], x -> 0]];
getGrowth[trunc@Last@#] & /@ pairs

Notebook

Motivation: getting a sense of asymptotics of Laplace transform for various random variables, which can be inferred from behavior of their CDF around 0

Answer 1

You can use Group and AsymptoticEqual.

exprs = getGrowth[trunc@Last@#] & /@ pairs;
Gather[exprs, AsymptoticEqual[#1, #2, x -> 0, Direction -> "FromAbove"] &]
(* {
 {3 x^2, x^2/18, E^(-((-2 + x)^2/(8 x))) Sqrt[2/π] Sqrt[x] - x^2/16 - (E x^2)/16}, 
 {Sqrt[x]/Sqrt[2], (2 Sqrt[x])/π, Sqrt[x], (2 Sqrt[x])/π, Sqrt[2/π] Sqrt[x]}, 
 {(9 x^3)/2, (4 x^3)/3}, 
 {x, x/2, x/(-1 + E), x, (4 x)/π, (2 x)/π, Sqrt[2/π] x, (2 x)/π, Sqrt[2/π] x, x}, 
 {0, 0}, 
 {E^(-(1/x^2))}, 
 {(3 E^(-3/x))/x}, 
 {-(E^(-(1/2) Log[x]^2)/(Sqrt[2 π] Log[x]))}
} *)

Answer 2

Your code produces a list

{3 x^2, Sqrt[x]/Sqrt[2], (2 Sqrt[x])/π, (9 x^3)/2, x, 0, x/2, x/(-1 + E),
E^(-(1/x^2)), (4 x^3)/3, x^2/18, (3 E^(-3/x))/x, x, Sqrt[x], (2 Sqrt[x])/π, 
(4 x)/π, -(E^(-(1/2) Log[x]^2)/(Sqrt[2 π] Log[x])), ( 2 x)/π, 0,
 Sqrt[2/π] x, (2 x)/π, Sqrt[2/π] Sqrt[x], Sqrt[2/π] x, x, 
 E^(-((-2 + x)^2/(8 x))) Sqrt[2/π] Sqrt[x] - x^2/16 - (E x^2)/16}

{3 x^2, Sqrt[x]/Sqrt[2], (2 Sqrt[x])/π, (9 x^3)/2, x, 0, x/2, x/(-1 + E), E^(-(1/x^2)), (4 x^3)/3, x^2/18, (3 E^(-3/x))/x, x, Sqrt[x], (2 Sqrt[x])/π, (4 x)/π, -(E^(-(1/2) Log[x]^2)/(Sqrt[2 π] Log[x])), ( 2 x)/π, 0, Sqrt[2/π] x, (2 x)/π, Sqrt[2/π] Sqrt[x], Sqrt[2/π] x, x, E^(-((-2 + x)^2/(8 x))) Sqrt[2/π] Sqrt[x] - x^2/16 - (E x^2)/16}

Now

Sort[%, AsymptoticLess[#1, #2, x -> 0,Direction->"FromAbove"] &]

{0, 0, E^(-(1/x^2)), ( 3 E^(-3/x))/x, -(E^(-(1/2) Log[x]^2)/(Sqrt[2 \[Pi]] Log[x])), ( 4 x^3)/3, (9 x^3)/2, E^(-((-2 + x)^2/(8 x))) Sqrt[2/\[Pi]] Sqrt[x] - x^2/16 - (E x^2)/ 16, x^2/18, 3 x^2, x, Sqrt[2/\[Pi]] x, (2 x)/\[Pi], Sqrt[2/\[Pi]] x, (2 x)/\[Pi], (4 x)/\[Pi], x, x/(-1 + E), x/2, x, Sqrt[2/\[Pi]] Sqrt[x], (2 Sqrt[x])/\[Pi], Sqrt[x], ( 2 Sqrt[x])/\[Pi], Sqrt[x]/Sqrt[2]}

does the job.

Edit. Direction->"FromAbove" is added to produce the more accurate result. Thanks to @Domen.