Timeline for Advantages of the sequence definition of limits
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 21, 2015 at 12:09 | comment | added | Liviu Nicolaescu | Try proving $\lim_{x\to 0}(1+x)^{1/x}=e$ not relying on differential calculus or on sequences. | |
| Aug 31, 2012 at 13:20 | comment | added | user9072 | Oh, sorry, for missing the tongue-in-cheek nature of that part of the argument. This makes more sense then. By the way, I like your books on the subject. | |
| Aug 31, 2012 at 13:00 | comment | added | Christian Blatter | @quid: Of course the thing with the cardinality was meant to be a joke. The essential point is that bringing in sequences replaces a simple limit by an infinity of composite ("verschachtelte") limits, over which reigns an additional $\forall$. | |
| Aug 31, 2012 at 12:27 | comment | added | user9072 |
Not to insists too much on what is essentially tangential to your argument (in my mind), yet since you changed the way you write the cardinality: even if it now says $\aleph_1^{\aleph_0}$ , this is still $2^{\aleph_0}$, the cardinality of the continuum (except perhaps in some more exotic set theories), and you won't get the thing 'cheaper' than with considering continuum many things somewhere. I think there really is no argument to be made based on cardinalities. To insist on brining this aspect in, in my opinion takes away from an (otherwise) very reasonable argument.
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| Aug 31, 2012 at 8:59 | history | edited | Christian Blatter | CC BY-SA 3.0 |
added 81 characters in body
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| Aug 30, 2012 at 20:59 | comment | added | user9072 | The above being said, while I am not fully decided on the general matter, I am aware of the fact that some students have difficulty with the 'for each sequence' and so on, for instance just showing something for one specific choice of sequence and thinking they are done. It is just this cardinality argument that I cannot follow, not the general point of view. | |
| Aug 30, 2012 at 20:43 | comment | added | user9072 |
Sorry, but the argument on the cardinality of the set of sequences one needs to test makes no sense to me. Why don't you have a problem with having to test for each of the $2^{\aleph_0}$ (assuming this is the cardinailty of the continum) many values of $\varepsilon$ that something holds for the typically $2^{\aleph_0}$ many $x$ in some $\delta$-neighborhood?
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| Aug 30, 2012 at 20:36 | history | answered | Christian Blatter | CC BY-SA 3.0 |