# Using the Squeezing Theorem of Limits in Mathematica

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.

• 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. – user64494 Jun 2 '18 at 17:23
• @user64494 But Reduce[ForAll[\[Alpha], \[Alpha] >= 0, 1 <= \[Rho] && \[Rho] <= 1 + \[Alpha]]] works just fine... – Henrik Schumacher Jun 2 '18 at 17:55
• @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. – user64494 Jun 2 '18 at 18:07

Limit[#, α -> 0] & /@ (1 <= ρ <= 1 + α)

• Yes, but this is a little weird to me! Why the Limit doesn't use its assumptions? What is wrong with my implementation? – H. R. Jun 2 '18 at 16:56
• @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. – AccidentalFourierTransform Jun 2 '18 at 17:07
• @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! – H. R. Jun 2 '18 at 17:20
• @H.R. Right, ρ lies in an interval with α-dependent interval boundaries. What AccidentalFourierTransform does is applying the limit to the interval boundaries. – Henrik Schumacher Jun 2 '18 at 17:54