1
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I wanted to compute the following limit

Limit[ρ, {α -> 0}, Direction -> "FromAbove", Assumptions -> 1 <= ρ <= 1 + α]

clearly this has the answer $\alpha$ by using the squeezing theorem. However, I don't know how to get the right answer from Mathematica.

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  • $\begingroup$ I think this topic is too hard for current CASes. For example, Limit[[Rho][[Alpha]], [Alpha] -> 0, Direction -> "FromAbove", Assumptions -> ForAll[[Alpha], [Alpha] >= 0, 1 <= [Rho][[Alpha]] && [Rho][[Alpha]] <= 1 + [Alpha]]] fails. $\endgroup$ – user64494 Jun 2 '18 at 17:23
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    $\begingroup$ @user64494 But Reduce[ForAll[\[Alpha], \[Alpha] >= 0, 1 <= \[Rho] && \[Rho] <= 1 + \[Alpha]]] works just fine... $\endgroup$ – Henrik Schumacher Jun 2 '18 at 17:55
  • $\begingroup$ @Henrik Schumakher: So what? The question is unclearly formulated. The command Limit[[Rho], [Alpha] -> 0, Direction -> "FromAbove", Assumptions -> ForAll[[Alpha], [Alpha] >= 0, 1 <= [Rho] && [Rho] <= 1 + [Alpha]]] produces [Rho] instead of 1. $\endgroup$ – user64494 Jun 2 '18 at 18:07
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Is this what you want?

Limit[#, α -> 0] & /@ (1 <= ρ <= 1 + α)
Reduce[%]
(* ρ == 1 *)
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  • $\begingroup$ Yes, but this is a little weird to me! Why the Limit doesn't use its assumptions? What is wrong with my implementation? $\endgroup$ – H. R. Jun 2 '18 at 16:56
  • $\begingroup$ Is ρ treated as a function of α in your code? Also the condition should be ForAll[[Alpha], [Alpha] >= 0, 1 <= [Rho][[Alpha]] && [Rho][[Alpha]] <= 1 + [Alpha]] . $\endgroup$ – user64494 Jun 2 '18 at 17:03
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    $\begingroup$ @H.R. Limit[ρ, α -> 0, whatever options] will always return ρ, regardless of what the actual options are. The reason is that ρ is independent of α, and therefore the limit is trivial. If you want a non-trivial limit, you'd have to use ρ[α] instead of ρ, to let MMA know ρ is a function of α rather than a constant. MMA is a piece of software, not a person, so it doesn't interpret intentions. $\endgroup$ – AccidentalFourierTransform Jun 2 '18 at 17:07
  • $\begingroup$ @user64494: ρ is not a function of α it is some constant which always lies in the aforementioned interval. Also, your assumptions does not does not solve the issue! $\endgroup$ – H. R. Jun 2 '18 at 17:20
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    $\begingroup$ @H.R. Right, ρ lies in an interval with α-dependent interval boundaries. What AccidentalFourierTransform does is applying the limit to the interval boundaries. $\endgroup$ – Henrik Schumacher Jun 2 '18 at 17:54

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