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Bug introduced in version 11.1.0 and fixed in version 11.2.0


The infinite limit

Assuming[a>0&&b>0,
  Limit[((1+a)*b*x)/Sqrt[1+(1+a)*b*x*((1+a)*b*x+Sqrt[1+(1+a)^2*b^2*x^2])],
    x->Infinity]]

gives the correct result 1/Sqrt[2] on version 11.0.0 but gives the incorrect result 7/8 on version 11.1.0.

What exactly has changed and what are the pitfalls we need to be watching out for?

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  • $\begingroup$ I am getting the same thing with V11.1. $\endgroup$ – bobbym Mar 29 '17 at 8:21
  • $\begingroup$ I get the same with versions 11.1.0 and 11.0.1 on Windows 7 x64. $\endgroup$ – Alexey Popkov Mar 29 '17 at 8:32
  • $\begingroup$ Maybe it has something to do with assumptions. The limit Assuming[a>0, Limit[Sqrt[a^2+x^2]/Sqrt[1+(x*(x+Sqrt[a^2+x^2]))/a^2], x->0]] gives a on version 11.0.0 but gives Sqrt[a^2] on version 11.1.0. $\endgroup$ – Roman Mar 29 '17 at 9:46
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    $\begingroup$ Will be fixed in a future release. Apologies for any trouble this may cause. It was a consequence of some code churn involving Series with fractional powers, on branch cuts. $\endgroup$ – Daniel Lichtblau Mar 29 '17 at 16:43
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    $\begingroup$ @DanielLichtblau Despite this bug, I am very much happier with the way the new Series (in 11.1) handles branch cuts of logs and fractional powers. I look forward to further development of Series in future releases. $\endgroup$ – QuantumDot Apr 3 '17 at 17:36
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Works correctly again in version 11.2.0:

Assuming[a > 0 && b > 0, 
 Limit[((1 + a)*b*x)/
   Sqrt[1 + (1 + a)*b*x*((1 + a)*b*x + Sqrt[1 + (1 + a)^2*b^2*x^2])], x -> Infinity]]

(* 1/Sqrt[2] *)

$Version

(* "11.2.0 for Mac OS X x86 (64-bit) (September 11, 2017)" *)
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