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The behavior of Limit changed in Mma 11.2, and I think it is in some ways undesirable or possibly buggy. Consider

Limit[Sin[t Pi], t -> Infinity, Assumptions -> Element[t, Integers]]

This used to (11.1) produce a correct result and no longer does (11.3). Is this intentional or a bug? (More understandable but still slightly disappointing is that Limit no longer returns limiting sets of values. E.g., removing the assumptions from the above expression, it used to return an interval, while it now returns the less informative Indeterminate.)

I am aware than I can change to DiscreteLimit, which is a nice addition, or use other tricks (e.g., Sum of differences).

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    $\begingroup$ v11.2 on my Mac also evaluates to 0 so although DiscreteLimit was introduced with v11.2, the change observed was not fully implemented until v11.3 $\endgroup$
    – Bob Hanlon
    Commented Aug 31, 2018 at 14:36

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As the documentation told me, the interval arithmetic can be restored with

Limit[Sin[x Pi], x -> ∞, Method -> {"AllowIndeterminateOutput" -> False}]

Interval[{-1, 1}]

I guess that this change was made for performance reasons, speeding up Limit with the default options.

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  • $\begingroup$ While this is useful for my subsidiary question, it does not address my main question. Is it a bug to apparently ignore the Assumptions? $\endgroup$
    – Alan
    Commented Sep 1, 2018 at 13:27
  • $\begingroup$ Well, I know. There a pros and cons for calling that a bug. I am rather indifferent in that matter. $\endgroup$ Commented Sep 1, 2018 at 13:32
  • $\begingroup$ What are the cons? Is there a motivation in the changes for the new behavior, which unlike the old behavior fails to appropriately use this assumption? (I'm not seeing it.) Just to be clear, it fails even if you change the Method as above. $\endgroup$
    – Alan
    Commented Sep 1, 2018 at 18:43
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    $\begingroup$ I submitted a report and received this answer: "The assumptions in Limit and related functions are only supposed to apply to parameters and not the limit variable." $\endgroup$
    – Alan
    Commented Sep 7, 2018 at 18:44

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