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Bug introduced in 9.0.0 and fixed in 9.0.1


I'm trying to compute this limit:

Limit[E^-(r^2)^(2*n)*r*2*Pi, n -> Infinity]

But, when I run the code above, Mathematica shuts down. I get a warning from Windows stating that the Mathematica 9 Kernel has stopped working. Mathematica itself gives no warning or message. Why is that?

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  • $\begingroup$ Which version do you have? Mathematica 9 doesn't crash, although it also can't evaluate this limit. If it's a new crash, you should report this as a bug. $\endgroup$ – Oleksandr R. Nov 14 '15 at 18:04
  • $\begingroup$ @OleksandrR. What I get is a warning from windows stating that the MAthematica 9 Kernel has stopped working... Mathematica gives no warning or message $\endgroup$ – An old man in the sea. Nov 14 '15 at 18:06
  • $\begingroup$ Are you using 9.0.0? 9.0.1 doesn't seem to have this problem. $\endgroup$ – Oleksandr R. Nov 14 '15 at 19:01
  • $\begingroup$ @OleksandrR. yes, 9.0.0 $\endgroup$ – An old man in the sea. Nov 14 '15 at 20:16
  • $\begingroup$ I've tagged it as a bug specific to version 9.0.0, because neither 8.0.4 nor 9.0.1 (the previous and next versions, respectively) have the same problem. I suggest updating Mathematica, if you can. $\endgroup$ – Oleksandr R. Nov 14 '15 at 22:43
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With

 $Version
 (* 10.3.0 for Microsoft Windows (64-bit) (October 9, 2015) *)

Mathemeatica returns a limit that depends on r, as it should.

Plot[Limit[E^-(r^2)^(2*n)*r*2*Pi, n -> Infinity], {r, -2, 2}, AxesLabel -> {r, Lim}]

enter image description here

Addendum

Although Limit returns unevaluated for arbitrary r, it can be made to produce useful results by making assumptions on r. For instance,

Piecewise[{Assuming[{#}, Limit[E^-(r^2)^(2*n)*r*2*Pi, n -> Infinity]], #} & /@ 
    {r < -1, r == -1, -1 < r < 1, r == 1, r > 1}]
(* Piecewise[{{0, r < -1}, {(-2*Pi)/E, r == -1}, {2*Pi*r, -1 < r < 1}, 
   {(2*Pi)/E, r == 1}}, 0] *)

which reproduces the plot above, apart from the discontinuities at r = -1 and r = 1.

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  • $\begingroup$ 9.0.1 can also plot this, although it can't evaluate the limit for arbitrary r. $\endgroup$ – Oleksandr R. Nov 14 '15 at 19:02
  • $\begingroup$ @OleksandrR. Sorry for my disjointed response a few hours ago. My plane was about to take off. Indeed, Limit also returns unevaluated for arbitrary r with Mathematica 10.3. But, putting assumptions on it produces useful results, as I just added to my answer. $\endgroup$ – bbgodfrey Nov 14 '15 at 21:50

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