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I tried to compute the following limit:

In[12]:= Limit[-A/2 x - 2 x, x -> +Infinity]

Out[12]= DirectedInfinity[-2 - A/2]

Does the output tell me that the software cannot evaluate the sign because it does not know the sign of A? And does it tell me that the solution is for sure plus or minus infinity?

Thank you in advance.

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  • $\begingroup$ DirectedInfinity is not a math notion, but an invention of Mathematica developers only. $\endgroup$
    – user64494
    Jul 27, 2022 at 6:21

1 Answer 1

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Does the output tell me that the software cannot evaluate the sign because it does not know the sign of A?

In a way, yes. But it does not even assume A to be real. What if A==I?

And does it tell me that the solution is for sure plus or minus infinity?

In a way, yes. But this is not generally correct. If A==4, then the answer is zero.

BTW newer versions give the answer as

(-2 - A/2) ∞

Also, in newer versions of Limit can do this:

Limit[-A/2 x - 2 x, x -> +Infinity, GenerateConditions -> True]

(* ConditionalExpression[(-2 - A/2) ∞, A != -4] *)

Finally, note that Infinity is also just DirectedInfinity in disguise.

Infinity//FullForm
(* DirectedInfinity[1] *)

-Infinity//FullForm
(* DirectedInfinity[-1] *)
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  • $\begingroup$ Thank you a lot @Szabolcs. I have a question for you: A != -4 means that the limit constant*infinity makes sense because constant=/=0? $\endgroup$ May 12, 2019 at 19:51
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    $\begingroup$ @GennaroArguzzi Yes, it is the reason why the condition is needed. Unfortunately, in your version of Mathematica (11.2 or earlier?) Limit does not have this feature. $\endgroup$
    – Szabolcs
    May 12, 2019 at 19:54

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