Tuesday, 14 July 2015

calculus - Is a limit a formalized infinitesimal?



From what I understand after thinking about this, delta epsilon really seems to formalize the notion of an infinitesimal. The constraint 0<|δc| combined with the fact that there is no real number that doesn't satisfy the delta-epsilon definition, then don't these two combined facts mean that an infinitesimal exists just not within the reals? In other words, there is a value that is greater than zero that is smaller than all possible real numbers (from my understanding), and the only value this could be is an infinitesimal. But I know that infinitesimals don't exist within the realm of real analysis. So where is my thinking wrong?


Answer



The opposition "limit versus infinitesimal" is a bit of a false opposition because limits are present both in the A-track approach working with an Archimedean continuum (the real numbers), and in the B-track approach working with a Bernoullian continuum (i.e., an infinitesimal-enriched one).



In the A-track, limit is defined via epsilon-delta definitions.




In the B-track, limit is defined in a more straightforward way using infinitesimals.



For example, lim can be defined simply as the standard part of f(\alpha) where \alpha\not=0 is infinitesimal.



Both epsilon-delta techniques and infinitesimals provide rigorous ways of handling the calculus. The relation between them can be stated as follows. The epsilon-delta techniques and definitions are a paraphrase of infinitesimal techniques and definitions.


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