Limits of functions in SMath.

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Best regards.
Alvaro.

Yes, I think that too. A note about the differences where Maxima assumes a positive for the result. Notice that in Maxima ∞ means +∞, but the idea in the SMath lim function is that ∞ means "unsigned infinitum", which is the true equivalent for 1/0, when 1/0+ = +∞ and 1/0- = -∞.



Let me explain.

What's the sense for distinguish the cases 0, 0+, 0- if you don't do the same thing with ∞? This is: if 1/x -> ∞ for x -> 0, why 1/x -> ∞ for x -> 0+, that's wrong: ∞ = 1/0 but ∞ = 1/0+. There are 3 infinity's in the real line: plus, minus and unsigned, which means infinity for both sides at the same time, like what happens with 1/x at x = 0. And because Maxima have two infinity's, not three, is why can't calculate the last lim(x*sin(x),x,∞)=und which is not +∞, nor -∞ but "∞", this is, unsigned ∞ which grows to infinity for "both sides": to +∞ and -∞ at the same time, like when 1/x decrease to zero for 0+ and 0- at the same time.



So Maxima have troubles for don't recognize the unsigned infinity: make continuous assumptions about the sign of the constants or directly can't give an answer for very simple examples. The infinitum in the lim function is an unsigned infinitum. Notice that in the complex plane the situation is more chaotic, you can grow to infinity in many distinct ways +∞ -∞*i, ∞ - ∞*i (unsigned in real axis, negative in the imaginary) ... etc.

Best regards.
Alvaro.

Yes, I wrote that for comparing with the SMath default's assumptions for the "infinity's algebra" (for calling that with something)





Yes, you're absolutely right. Sorry, I don't know about what I could be thinking at that moment.



You're right again about it is practical in a CAS or not. But it is not related about the unary operators + or -, it is related with the topological concept of path in a domain. For reals you have three paths for go to zero: by left, by right or by both sides at the same time. All CAS can handle that very well, but have troubles handling the inversion of that: 1/0+ = +∞ , 1/0- = -∞ but 1/0 = ∞ unsigned. For example, I don't see what the sense of not recognize this limit:



Also I don't think that assign that result to infinity could surprise anyone. In the classic literature also you have unsigned infinity everywhere.

Rey Pastor, Análisis Matemático vol I, page 274 first introduce unsigned infinity limit for successions and +∞ -∞ as particular cases



Piskunov, Differential and Integral Calculus, page 53 take an entire page for demonstrate that a function converges for unsigned infinity and put that as a theorem.


That could be too much for real calculus, but I guess that it is the preamble for work with the complications of the topology of the complex plane in a further study of limits. So, I agree with you that it could be impractical to introduce it in SMath, but also what I see are unnecessary complications without it.

Best regards.
Alvaro.