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Yeah, I hadn't thought this fully through when posting, and was almost right more by accident than anything. To further show your point, if you actually take th
by codeflo 4y ago
Yeah, I hadn't thought this fully through when posting, and was almost right more by accident than anything. To further show your point, if you actually take the (real) limit, you get lim t->0 (x^t-1)/t = ln(x) using L'Hopital's rule, so ln(x) = (x^ε-1)/ε makes some sense in that regard.
- hansvm 4y agoYep. That's actually one of the main points of infinitesimal theories -- all the standard theorems have infinitesimal analogues and vice versa. Rounding to the closest real number in an infinitesimal theory is the same as taking a limit in real theory. It is definitely nice to have multiple routes to the same answer though to help verify that some stupid mistake didn't infect the results.