# Limit in Infinite

My code:

a = 1.26957; k = 0.607248;
tailF[x_] := Exp[-a*x^k];
m = 1; b = 3;
f[x_] := (m*(2*π*b*x^3)^(-0.5))*Exp[-((x - m)^2)/(2*b*x)];
mesi = N[Integrate[x*f[x], {x, 0, Infinity}]];
diaspora = N[Integrate[x^2*f[x], {x, 0, Infinity}] - (Integrate[x*f[x], {x, 0, Infinity}])^2];
ouraF[x_] := Integrate[f[t], {t, x, Infinity}];
N[Limit[ouraF[x]/tailF[x], {x, Infinity}]]


I m making something wrong when i calculate the Limit. The denominator have a Integrate and their is the problem how i can express it!

• Inexact numbers like 0.607248 usually don't play nice with Limit[], so if you can get exact expressions for those, it would be useful (or just use Rationalize[]). Commented Jan 22, 2020 at 6:23

Here's what works for me

ClearAll[a, k, f, tailF, ouraF]

tailF[x_] := Exp[-a*x^k]

m = 1; b = 3;
f[x_] := Sqrt[m*2*\[Pi]*b*x^3]*Exp[-((x - m)^2)/(2*b*x)]

ouraF[x_] := Integrate[f[t], {t, x, Infinity}, Assumptions -> x > 0]

Limit[ouraF[x]/tailF[x], x -> Infinity, Assumptions -> {0 < a, 0 < k < 1}]

(*  0  *)


The changes are (1) use Sqrt instead of ^(0.5), (2) prefer symbolic integration, where possible, (3) apply assumptions when Integrate has a symbolic limit and (4) assumptions can be used with Limit, too.