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?)

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

1 Answer 1


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]}]

  • $\begingroup$ 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. $\endgroup$ Jan 5, 2017 at 16:26
  • $\begingroup$ @ManuelEberl. Wolfram Technical Support can only fix it. $\endgroup$ 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) $\endgroup$ Jan 5, 2017 at 21:28
  • 1
    $\begingroup$ I'm have MMA version 10.2 and 11.0 both can't find solution. $\endgroup$ Jan 5, 2017 at 21:37
  • 1
    $\begingroup$ Neither can 10.4.1. $\endgroup$
    – corey979
    Jan 5, 2017 at 21:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.