Let epsilon < 0.
I feel I should understand it, but it's just outside of my reach. It's now 10 years after university.
I don't think you can use the x0 plus minus delta in the bracket (or anywhere), because then the function that's 1 on the rationals and 0 on the irrationals is continuous, because no matter what positive number epsilon is, you can pick delta=7 and x0 plus minus delta is exactly as rational as x0 is so the distance to L is zero, so under epsilon.
You have to say that
whenever |x - 0x|<delta,
|f(x) - L|<epsilon.But I think this is one of my favourite memes.
unless f(x0 ± δ) is some kind of funky shorthand for the set { f(x) : x ∈ ℝ, | x - x0 | < δ }. in that case, the definition would be “correct”.
it’s much more likely that it’s a typo, but analysts have been known to cook up some pretty bizarre notation from time to time, so it’s not totally out of the question.
... That's enough real analysis for me today. Or ever, really.
Yeah
i still feel like this whole ε-δ thing could have been avoided if we had just put more effort into the “infinitesimals” approach, which is a bit more intuitive anyways.
but on the other hand, you need a lot of heavy tools to make infinitesimals work in a rigorous setting, and shortcuts can be nice sometimes
Infinitesimal approach is often more convoluted when you perform various operations, like exponentials.
Instead, epsilon-delta can be encapsulated as a ball business, then later to inverse image check for topology.
i think the ε-δ approach leads to way more cumbersome and long proofs, and it leads to a good amount of separation between the “idea being proved” and the proof itself.
it’s especially rough when you’re chasing around multiple “limit variables” that depend on different things. i still have flashbacks to my second measure theory course where we would spend an entire two hour lecture on one theorem, chasing around ε and η throughout different parts of the proof.
best to nip it in the bud id say
Calculus, Motherfucker! Do you speak it?!
Not a mathematician, but I'm pretty sure this isn't necessarily true. What if L is -1 and f(x) = x^2? Also I think your function has to be continuous.
You're right on all three counts. It's not always true, f(x0) has to be L, and the function has to be continuous.
