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Mathematica 10.1.0 fails to compute the following limit:

Limit[Log[X]^(3/2)*(-1 - 1/Sqrt[Log[X]] + (1 + 2/Log[X]^2)^p*
        (1 + 1/Sqrt[Log[X/2 + X/Log[X]^2]])), 
  X -> Infinity, Assumptions -> {Element[p, Reals]}]

After ten seconds or so, it simply returns the unevaluated expression even though the limit is $\frac{1}{2}\ln 2$ irrespective of the value of $p$. I'm fairly sure that this is the case because:

  • I've proven it in a proof assistant (Isabelle/HOL)
  • Maple outputs the correct result
  • Mathematica outputs the correct result if any concrete value is substituted for $p$

Is this some kind of fundamental restriction of Mathematica regarding free variables in expressions or am I missing something here? (Someone in the ##mathematica IRC channel claimed it worked fine in his Mathematica version, so perhaps the version is relevant?)

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  • $\begingroup$ I too have noticed problems with simple limits in recent versions of Mathematica $\endgroup$
    – BeauGeste
    Commented Jan 5, 2017 at 22:59

1 Answer 1

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I don't know why Mathematica has problems,but if substitute X->Exp[x] then:

sol = FullSimplify[Log[X]^(3/2)*(-1 - 
1/Sqrt[Log[X]] + (1 + 2/Log[X]^2)^p*
(1 + 1/Sqrt[Log[X/2 + X/Log[X]^2]])) /. X -> Exp[x], Assumptions -> {Element[x, Reals]}]

(* x^(3/2) (-1 - 1/Sqrt[x] + (1 + 2/x^2)^p (1 + 1/Sqrt[x + Log[1/2 + 1/x^2]])) *)

Limit[sol, x -> Infinity, Assumptions -> {Element[p, Reals]}]

(*Log[2]/2*)
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  • $\begingroup$ That's an interesting observation, thanks. I am, however, interested not in an ad-hoc way to make this particular example go through but rather in understanding what is going wrong here and how to fix it in general. $\endgroup$ Commented Jan 5, 2017 at 16:26
  • $\begingroup$ @ManuelEberl. Wolfram Technical Support can only fix it. $\endgroup$ Commented Jan 5, 2017 at 17:00
  • $\begingroup$ Well, by ‘fix’ I meant finding e.g. a different set of configuration options. I don't know Mathematica very well and thought I might have missed something. (Especially because someone on IRC claimed that the examples works fine with his version of Mathematica) $\endgroup$ Commented Jan 5, 2017 at 21:28
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    $\begingroup$ I'm have MMA version 10.2 and 11.0 both can't find solution. $\endgroup$ Commented Jan 5, 2017 at 21:37
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    $\begingroup$ Neither can 10.4.1. $\endgroup$
    – corey979
    Commented Jan 5, 2017 at 21:58

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