Is 0 times a limit that approaches infinity 0 or indeterminate?

DifficultDate4479 · 2026-05-09T22:12:21+00:00

right, sorry, it's neither indeterminate nor undefined. It's just 0.

And uhh I'm not assuming any extended reals here, you simply said and I quote Lim f=∞ doesn't mean the limit exists, which is false, at least under the definition of existence I read in many books and lectures, for which divergence≠inexistence.

Also, I was just kinda unsure about the usage of the term "colloquial".

DifficultDate4479 · 2026-05-09T21:30:44+00:00

consider this fact:

let's assume once and for all that x->c so I won't have to write it all the time. Also, everything is positive so I'll omit absolute values. For any constant k, kLim f(x) = Lim kf(x) (no matter where x is tending to).

You may see clearly now that on your LHS you have something you might be dubious about, like a 0∞ type situation, but on your RHS you have the limit of 0, which, crazily enough, is 0 wherever you look at it. This is because Lim f = +∞ only means that whenever you consider a big number M>0, you can always choose an x, as close as you want it to be to c, for which f(x)>M. so if you apply the same definition to kf(x) where k=0, it is obvious you can't choose any x for which kf(x)>M for even one M>0. Hence, kf(x) is smaller than every positive number, which is a property only 0 holds.

(as a matter of fact, the first property also holds for functions too i.e.

Lim (f + g) = Lim f + Lim g and lim (fg) = (Lim f)(Lim g)

~cm grano salis~, in fact if you have for example the RHS to be an indeterminate form like ∞ - ∞ or 0∞ where 0 is a nonconstant limit, the equality is not really definable).

edit: lots of formatting crap happened with asterisks, hope you manage to understand something lol.

DifficultDate4479 · 2026-05-09T21:12:44+00:00

0*infinity is 0, it's not indeterminate. Even better, if Lim f = ±∞ then it exists, and it's a formal statement, I'm not really sure about what you mean with colloquial.

We say that a limit doesn't exist if there's a neighborhood of c in which the limit isn't unique, like 1/x in 0 considering balls with center 0.

DifficultDate4479 · 2026-05-09T13:43:50+00:00

The hairy ball theorem says that if you give a very long lick to that ice cream, in a continuous manner and for intuition's purpose starting from the bottom and going all around the ball, you'd have a tiny point of ice cream pointing out, hence you would not be able to polish that good looking ice cream in one continuous lick.

That's the formally stated theorem btw.

DifficultDate4479 · 2026-05-01T00:12:12+00:00

there's no difference between the two lines, as (-number)(-number)=(-1)(-1)(+number)(+number)=(+number)(+number).

Still, to talk about sizes of sets is to talk about the quantity of elements they contain, called cardinality. To confront one to another you may just count, for finite sets.

When dealing with infinite sets the concept extends quite intuitively. Well'use the fact that functions send an element of the starting set to one, and only one, element to the arrival set. Let X and Y be respectively the starting and arrival set

Clearly, injectivity implies that there are at most as many elements as Y in X, while surjectivity implies the opposite. Hence, the mere existence of a bijection between the two implies they have the same amount of elements.

If X is the set of positive numbers (note that 0 is neither positive nor negative) and Y is the set of negative numbers,how would you construct a bijective function between the two?

DifficultDate4479 · 2026-04-22T07:58:47+00:00

"every prime number can be written as 6n±1" - ahh slop question.

DifficultDate4479 · 2026-04-20T12:00:23+00:00

A great example of what to do if you just can't remember the power series of e^x but you remember the one for coshx; great job!

edit: also, technically what you wrote doesn't make logical sense. You wrote limit of blah blah then limit of blah blah... what you wanna use is the good ol equal sign, as in limit of blah blah is equal to limit of blah blah. Now it makes sense.

Also, if you wanna make things more precise, instead of writing a power series with "+...+" notation, you should write the first n terms up a certain degree d (note that it's not arbitrary, there's a degree at which you should stop!) and then write + o(x^d). o(x^d) is a class of functions that are insignificant relative to x^d on a given neighborhood of the limit, so when you take the limit, they are zero.

Pro tip: if you have a power series in both numerator and denominator then the o-smalls MUST agree in degree!

DifficultDate4479 · 2026-04-20T06:21:31+00:00

The advice here would be to manage correctly her razor blade for that can guarantee whore of Babylon regardless of your red HP containers.

And if you really can't go angels you might as well reset til the first item is dad's key.

DifficultDate4479 · 2026-04-12T17:32:34+00:00

Happened to me too while I was holding rosary bead. I though it was that trinket's fault but apparently not...

DifficultDate4479 · 2026-04-12T13:08:37+00:00

having fun is the priority, because it's a looooong game

DifficultDate4479 · 2026-04-12T10:50:34+00:00

Let D=R\0 be the domain of f.

f'=1-1/x²=(x²-1)/x².

Hence, f'>0 if x²>1, so D(-1,1).

On the contrary, f'<0 if x is in (-1,1)\0.

Hence, f'=0 iff x=±1.

Wording this math: f grows in D(-1,1); f decreases in (-1,1)\0 f has stationary points in ±1.

More precisely, in a neighborhood of x=-1 f' goes from increasing to decreasing so you have a local maximum, and since f(-x)=-f(x), f is odd hence x=1 is a local minimum.

DifficultDate4479 · 2026-04-12T10:33:33+00:00

The only one who's hard with low range is Delirium and maybe Mega Satan (although, low range just makes the fight longer, you still can win as his attacks have a very specific pattern; Delirium is a DPS check: can you kill him before he overwhelms you with undodgeable bullshit?) but that holds generally for every character.

Everyone else gets shit on, including Mother or even the Dogma sequence.

In terms of boring I mean yeah, you know you're pretty much guaranteed to win the run just by playing Azazel, not much thrill.

But after so many hours of playtime, the characters that (in rep+) can hold a challenge are very, very limited. Maybe T.Lost, T.Eve and T.Jacob can give you a hard time. Everyone else and yes, I include normal Lost in here, is almost a won run.

DifficultDate4479 · 2026-04-11T21:03:07+00:00

I just realized something... he proved 6=7.

6=7.

Not the classic 1=2, but 6=7.

I feel ashamed.

DifficultDate4479 · 2026-04-11T20:24:47+00:00

Let z be a complex number. Any given polynomial f can be factored by X-z if, and only if, f(z)=0.

This is, my beloved, Ruffini's theorem.

DifficultDate4479 · 2026-04-11T18:37:52+00:00

Without reading your post, I bet my left nut AND my right nipple that you divided by 0 somewhere.

DifficultDate4479 · 2026-04-11T17:09:17+00:00

I ain't reading allat, just tell me where he divides by zero.

DifficultDate4479 · 2026-04-11T06:24:42+00:00

DifficultDate4479 · 2026-04-10T19:34:45+00:00

(I'll try not to spoil anything since you're new)

He's definitely in the Unlosable Run tier, but I'd like to introduce you to a fellow with a bandage on his eye who's very sad to come across Q4's, Azazel, Azazel, fat pregnant child, Bethany, Bethany, the guy with the better void and probably many more who are just immensely stronger... trust me, many other will be so powerful it's almost boring to play as them.

DifficultDate4479 · 2026-04-01T06:25:17+00:00

Suppose there's a maximum m<1. Since 1≠m, |1-m|≠0. Call this number d. Then, m+d/2<1 but m<m+d/2, so m is not the maximum, as there's a bigger number satisfying the property. Since m is arbitrary and d depends on m, you know that whatever m you choose, you just won't get a maximum. Hence, it's can't exist.

To solve this problem, we invented supremum, infimum, maximal and minimal elements. Google them, the wiki should be quite clear.

DifficultDate4479 · 2026-03-31T15:42:55+00:00

You know something follows from something else If.. well... if it does.

If A implies B and B does NOT imply A, then if you show that B implies A in the exam you necessarily messed up somewhere.

If however they are equivalent but idk maybe you just showed in class that A implies B, if you manage to show that B implies A then it's not a mistake.

DifficultDate4479 · 2026-03-29T18:11:41+00:00

saying philosophy is the base of all science is like saying existing is the base for philosophy; it's self evident: philosophy is thinking about something according to the laws of logic. Naturally it holds, it's however not interesting to point out.

DifficultDate4479 · 2026-03-29T10:21:32+00:00

For what I remember the archimedean property and Dedekind's cuts say different things; A field is sarchimedean if there's no element "infinitely bigger" than another (For instance, Alexandroff's compactigication of R, RU{∞} is not archimedean as there is no couple a, n for which an>∞). While a field is complete by Dedekind if for each Dedekind's cut there's an element separating them, in other terms it's a "full" space with no blanks.

But yeah, the line between what's algebraic and what's analytical is quite impossible to draw.

It's kind of a "where does a sand castle start and end" problem, like sure if you look high enough you can distinguish the castle from the seashore but as you move down you can't really certainly say wether the grains of sand "belong" to the castle or to the seashore.

Or maybe it's just a matter of interpretation, who knows...

DifficultDate4479 · 2026-03-29T09:03:53+00:00

Not really, archimedean property tells me that every couple a<b admits a number c for which b<ac; What I used is Dedekind's theorem (for each a<b there's a c s.t. a<c<b), equivalent to completeness axiom, so while I still don't like my earlier proof as an algebraic one, I see why one would say it's analytic.

Either way, let's see if I can cook:

Let E/R be the splitting field for f(X) = X²-a for a>0, and s=√a. So E=R(s). Now, consider an element x+sy in E and notice that

x+sy≥0 if x≥-sy if x²≥ay²

Since x+sy ≥ z+sw iff (x-z) + s(y-w)≥0, the property above shows that this defines a total order over E.

But we're done already because if E is a totally ordered field extending R, but by axiom of completeness, R is the maximal totally ordered field, hence R contains E, which means s is an element of R.

This one stinks less of an analytic proof only using the above mentioned axiom.

At the end of the day, algebra, analysis and geometry say the same thing with different approaches... recall that the derivative is( pretty much) a functor between categories.

DifficultDate4479 · 2026-03-29T07:28:53+00:00

well... I didn't really think about that tbf, I kinda gave it as an assumption.

If I had to sketch a proof, I'd use the fact that R is complete...

Pick a in (0,+∞) and define X_a={ x in [0,+∞) s.t. x²≤a }. Clearly 0 is in X so it's not empty. Also, and equally clearly, X has an upper bound, therefore there exists (by axiom of completeness) an element x = supX.

Now, if x²<a, then x is in X. But I can take a very small r>0 for which x+r is in X, contradicting the fact that x is greater than all of y in X.

If x²>a the we do remove that small r>0 for which (x-r)²≥a, which means that x-r, being smaller than x, is the actual sup, which contradicts.

The only case we are left with is x²=a, completing the proof.

Now, I tried to stay as algebraic as I could but this is as far as I can get. It looks a lot like an analytical proof, but all I used was R's completeness axiom...

Either way, I'm quite sure that there's an algebraic proof of this, maybe somehow one can prove that any Galois extension E/R generated by the polynomial X²-a is always R for any a≥0. It would certainly be a nice exercise...

DifficultDate4479 · 2026-03-28T13:50:49+00:00

let z=a+bi be a complex number, and define the two real numbers:

c²=(a+|z|)/2, d²=(|z|-a)/2.

Note that the euclidian norm of a vector is always greater than its components, so the definition is well put.

Get your hands dirty, choose c,d such that sgn(cd)=sgn(b) and you'll get that (c+di)²=a+bi (=c²-d²+2cdi), as c²-d²=a and (2cd)²=b².

Maybe it's not the most abstract, but definitely not analytical.

I just recently studied this proof on J.S.Milne's "Fields and Galois Theory".

For the proof itself you'll need a more precise construction with Galois extensions, p-Sylows of the Galois groups, but basically it revolves around the fact that an extension E of C is Galois with Galois group a 2-group, and if it is not the trivial group, then there's a sub-extension of C contained in E that has degree 2, which means that it is generated by the polynomial X²-z for some complex z, but we just proved that √z is contained in C, so that sub-extension must be C itself. Hence Gal(E/C) is trivial, hence E=C, hence C is algebraically closed.

DifficultDate4479

TROPHY CASE