1 ^ ∞, It's Not What You Think 

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What we learn in class: 2² = 4 Whats on the homework: 2³ + 3 = ? The exam:

As a physicist, I'm going to forget I ever saw this so I can still do... well a surprising amount of physics that relies on 1^infty = 1 lol.

I rely on the ‘come on man’ theorem on questions like this. The way it works is if there’s a difficult question in math you assert your opinion without backing it up in any way. And if someone challenges it you just say ‘come on man.’

The only sensible interpretation is that 1^oo = lim n->oo 1^n and that is 1.

Based on my calculations, this video will be epic

1:46 The reason why you get different answers is because both the bases are different numbers. They are not equal to 1, just two different numbers that are really really close to 1. An exact 1 to the power of infinity is equal to one.

What I don't understand about analysis like this is the whole idea is to get mathematics consistent, correct? So we, like you said, can choose 1∞ to be whatever we want as long as it's mathematically rigorous and consistent. So let's say 1^∞ is 1, is there any other (potentially seemingly unrelated) mathematics that breaks with this definition? Because if there isn't, I don't see any issue with defining it this way. The example used here where the limit as x -> ∞ of (1 + 1/x)^x = e might be a good counter-example to this point, but the issue I have here is that it doesn't really tend to exactly 1^∞, it's more tending to (1+)^∞, but I guess an argument could be made here that this should diverge to infinity, while it clearly doesn't - this is another issue though.

Okay, so before I start giving mathematical explanations, I need to get this out of the way: mathematics educators and communicators seriously need to stop talking about "indeterminate forms" as a legitimate mathematical concept, because I think they are doing a great disservice to people who are trying to get a deeper understanding of the mathematics of calculus and arithmetic. *There is no such a thing as an 'indeterminate form.'* No, in mathematics, there are two, and only two classes of expressions: those which are well-defined, and those which are undefined, or ill-defined (I take "undefined" and "ill-defined" to mean the same thing, as there ultimately is no coherent way for the two to be meaningfully distinguished). An expression like 2^3 is well-defined. An expression like 1^∞ is not. The latter is in the exact same category as 1/0, ln(0), 0/0, 0•∞, and so on. Please, let us start calling these things for what they are: undefined expressions, because that is all that they are. An expression is just a string of symbols. We can take symbols to either be fundamental in signficance (i.e., the symbol by itself means something), or we can take a few symbols that, when combined into a string, that string has fundamental significance (e.g., the string "sin," which is comprised of the symbols 's,' 'i,' 'n,' symbols which have no significance of their own, but when combined into the string "sin," they form a string with some actual mathematical significance). When you combine multiple fundamental, 'atomic' strings into an expression, the significance of this composite expression depends, in some way, on the significance of the fundamental symbols, and how that significance emerges from the combination is dictated by rules of notation. The expression 1^∞ is an expression with three atoms: the atom "1," the atom "^," and the atom "∞." We need to discuss how these individual symbols are defined, before we can say anything about 1^∞. What is the symbol 1 defined as? Well, we are secretly working with some fundamental type of mathematical structure obeying certain axioms formulated in some type of logic. This structure is called the ordered field of real numbers, and these are ultimately just some abstract objects that do nothing except satisfy certain properties we call 'axioms,' properties that are meant to be understood rules for how these objects interact, and these objects we take as being primitive or fundamental in this context: they are not defined in terms of anything else, they kind of just "are" for the sake of being alone. These real numbers satisfy one axiom, that there exists some y, such that for all x, x•y = y•x = x. This axiom is one of the defining properties of •, a binary operation, and when we use the other axioms, we can prove formally that this y object satisfying the above property is unique. Since it is unique, we give it a name. The symbol "1" is that name. So, 1 is defined by the property that it uniquely satisfies: that 1•x = x•1 = 1, for all x. That is one symbol out of the way. Now, what is ^ defined as? It is a function, specifically a function on the Cartesian product between R+ and R, with R as the codomain. R stands for the ordered field of real numbers, the fundamental structure we are interested in studying, and R+ refers to those real numbers which satisfy 0 < x, they are called positive real numbers. ^ is defined by the property that x^1 = x, that if x < y, then (1 + 1)^x < (1 + 1)^y, and that x^(y + z) = (x^y)•(x^z). Notice how, in turn, this requires understanding the definitions of < and +, which are all defined by axioms themselves. Everyone here intuitively understands how these objects work, so I will not bother with every detail, but the point is, it matters to know where the foundations of the discussion lie. The one symbol here that is troublesome here is the symbol ∞. This symbol is technically well-defined: it is the maximum of the Dedekind-McNeille completion of the real numbers, which is a mathematical structure that extends the real numbers in a very specific way. However, the problem is that the way the symbol is defined, it does not for any arithmetic. There is no mathematically consistent way to extend to the domains of + and • as functions to include ∞ in the domain, and continue having well-defined expressions. We have tried, but because of the axioms these structures are defined by, it can be proven that there is no way to make it work. Things like ∞ - ∞ and 0•∞ will always cause problems. So, arithemetic expressions with ∞ are *undefined.* They are *undefined,* because the way mathematical definitions work makes them a very specific kind of thing. Mathematical definitions have to make reference to mathematical properties: to well-defined mathematical concepts, and we have to ensure that the definition does not secretly hide a contradiction within it (which makes it incoherent), and we have to make sure that whatever properties are being referenced are actually uniquely satisfied if the definition is meant to denote a particular object, and not an entire class of objects. Since the axioms make it impossible for ∞ to be in the domain of any suitable extensions of +, •, and ^, the objects that satisfy the properties that the expression 1^∞ would make a reference to simply do not exist. In other words, the expression does not refer to anything at all! It has no actual significance, and this what we mean when we say the expressions are undefined. Definitions in mathematics are not an arbitrary endeavor, like everyone likes to say. There is some degree of arbitrariety, yes, but only after you get past many walls of restrictions that must be satisfied. Now, what is the funny business with indeterminate forms? Why do people insist to interpret 1^∞ as being a "shorthand expression" for lim f(x)^g(x) (x -> p), specifically in the cases that lim f(x) (x -> p) = 1 and lim g(x) (x -> p) = ∞ (whatever this expression even is supposed to mean)? The problem here is that these two notations actually make reference to two completely unrelated mathematical topics that are different. We need to stop making an association and pretending these symbols are synonymous. Do we use the symbols floor(0) and lim floor (x -> 0) to mean the same thing? No, we do not, and the questions "what is floor(0)?" and "what is lim floor(-x^2) (x -> 0)?" have completely different answers. floor(0) = 0, while lim floor(-x^2) (x -> 0) = -1. Incidentally, lim floor(x^2) (x -> 0) = 0, which is not equal to -1. Does that mean floor(0) is "an indeterminate form"? Hmm??? I ask this in a completely serious fashion, just to highlight how, um, inaccurate (to put it politely) the notion of "indeterminate forms" is. No, floor(0) is not an indeterminate form. No one thinks of it as one. We all know that floor(0) = 0, and if I ask you "what is lim floo(x) (x -> 0)?," then you would, correctly say, that the expression is undefined/ill-defined. Alternatively, you would say that the limit being referenced by the expression does not actually exist: there is no mathematical object that we can call "the limit of floor(x), x -> 0," because the definition is not satisfied by any objects at all. So, despite this being the case, why are people still insisting on this indeterminate form concept? I do not understand it. The question "what is 1^∞?" is a completely different question from "if lim f (x -> p) = 1 and lim g (x - p) = ∞, then what is lim f(x)^g(x) (x -> p) equal to?," and the two questions have different answers. For the former question, the expression given is undefined. *Why* is it undefined? To answer that, you have to get into a discussion of Dedekind-McNeille completions as mathematical structures, and why their properties do not allow for + and • as operations. The latter question, though, has a simple answer: it depends on the properties of f and g as functions. Do you see how these are different questions? One question is about arithemetic, abstract algebra, and about mathematical structures. The other question is about functions of real numbers and their asymptotic behaviors. Completely unrelated topics. So, why does the education system insist in treating them as the same question and as the same topic? I have no idea why, but we need to stop doing it. And, yes, students do need to understand that knowing lim f (x -> p) = 1 and lim g (x -> p) = ∞ is *not* sufficient information to determine what lim f^g (x -> p) is. But using the language of "indeterminate forms" to teach this is unhelpful, and very misleading. Firstly, it gives people the impression that being "indeterminate" is somehow fundamentally different from "undefined." Secondly, it reinforces the already too common misconception that to evaluate limits, you just "plug in" and see what happens (not how limits work at all). Thirdly, it leads to people developing a bunch of mystical, crackpot nonsense like "ah, so this number/expression is equal to all numbers at once, which proves we are all one [insert more mysticism]." Fourthly, it is just an unnecessarily obfsucating, convoluted way of telling the student "we do not have sufficent information to answer the question." Fifthly, whenever we bring up limits to answer questions like the value of 1^∞, we do a disservice by actually failing to answer the question, and instead, answering a different, unrelated question.

Also you can approach this like the 0^0 value is approached: if we want a^m a^n = a^(m + n) to hold and we allow ourselves ∞ + ∞ = ∞, then we have a condition 1^∞ 1^∞ = 1^∞, i. e. this should be a multiplicative idempotent. If we want it to be a real or complex number then we already have just two options left: 1 and 0. Now, we can make a couple of arguments to discredit 0 as the potential value, accurately enough to leave 1 as valid. Voilà!

“But, is it?” **cues the VSauce Intro Music*

I think this is more a question of mathematical language than mathematical reasoning. If 1^x were a standalone function like sin x or ln x, setting 1^∞ to 1 would be fine. However, we're dealing with a two-input operator x^y here. That means you can approach the (projective) point x=1, y=∞ from infinitely many directions. Tracing the value of x^y on those approaches 1 or 0 or ∞, or even any value between those (needs curved approach): let a_n, b_n > 0 infinite sequences with a_n≠1, lim a_n = 1, lim b_n = b. Set x=a_n, y=(ln b_n)/(ln a_n) then x^y approaches b as n approaches infinity, while x approaches 1. If b≠1, |y| approaches infinity. The sign of y can be controlled by putting a_n on the same side of 1 as b is.

2:55 an interesting reference here could be mentioning that X^Y is set of all functions from Y to X, then say about Von Neumann's definition of natural numbers. Empty set is 0, while 1 = {0}, thus if we think about ∞ as the set Y with infinitely many elements, then 1^∞ can be interpret as {0}^Y, i.e. set of all functions from Y into {0}. However, it contains only one constant function that is always equal to 0 😄 The result is still not a number, but the same could be said about "1^∞", so it's an interesting interpretation of it 🙂

The limit of q ^ n when n goes to infinity depends on the value of q q = -1 undefined since it's an oscillating function like sin and cos -1 < q < 1 it's 0 q > 1 it's infinity q = 1 it's actually equal to 1 but only because it's the variable n which is part of Z Assuming the value is exactly 1 and not (1+) like in that example, it's undefined if the variable is a real number and 1 if the variable is from Z

I remember hearing this once and it will always be stuck in my head: Infinity is not a number, it’s a concept

This is one of the best math videos I have seen in a while! Very concise and informative. You have earned yourself a subscriber :)

I've seen so many videos about all these things lately, and what amazes me the most is that everybody in their channel talks like if they were the first ones ever trying to find an answer. All these things must be already defined by mathematicians long ago or they wouldn't be able to solve many problems.

Nice video, as always. I have a problem with the sentence "You take the base and multiply it by itself exponent times" (1:35). There are clearly 3 multiplications in the decomposition of 2 to the 4th. If you multiply the base by itself ONCE you get the base squared. I think a better sentence would be "You take 1 and multiply it by the base exponent times". This would also account for why any base to the 0th power is 1.

2:32 - that is really interesting how we have gone from something obvious to something impossible

Thanks! i will go straight to using this new definition without any context or explanation in my real analysis exam

Sorry, but this seems about as useful as medieval bishops arguing over how many angels would fit on the head of a pin.

I waited sooooo much for this video! Thanks a lot!

A video from Bri a day keeps ignorance away

Well, one way to approach this would be to take a number system where you do have infinite numbers and power functions, take some reasonable choice on what infinite number the infinity symbol should stand for (or alternatively, find out that for the number system of choice it doesn't matter), and enjoy the well defined result here. Two obvious choices here would be ordinal numbers and cardinal numbers. In ordinal numbers, one might choose the smallest infinite number, omega. Now by definition (well, I think there are several equivalent definitions, but this one is particularly convenient) a^omega = sup(a^n), n

X^infinite= 1 if X is absolute 1.But if X is trending to 1 then,X^infinite is not equal to 1.

1 (as the numerical constant 1) elevated to the whatever is 1, unless that "1" isn't really 1 aka some function approaching 1

What I learned is that my infinite bank account with $1 on it is never gonna increase.

"its not what you think" proceeds to say it is exactly that

the thing is 1 times 1 will always result in 1,no matter how many times you repeat the operation

In case anyone is wondering what does happen when you put one raised to infinity into a calculator, it goes like this: Casio fx-CG500: 1 TI-89 Titanium: 1 HP Prime: undef TI-nspire CX II: 1 with the message “Warning: 1^oo replaced by 1”

Great video and super interesting. Quick note 1^∞ is an indeterminate form meaning that we need to determine it.

Why is the example at 1:05 used as what the written form for the limit would be, instead of the one at the end? Also, why are you calling into question if the audience is sure of the definition at 3:13? That is what I imagine all exponents are, in reality. Am I misunderstanding something? Even in that example, it seems to me like it would just lead back to always being 1, instead of being undefined. Thank you for the video.

When dealing with infinity, which is not a number, the result you get depends on specifically how you pose the question. Thus, it is indeterminant. Your math example compelled me to think about the following problem. Consider. y= (1+ (1/x^z))^x As you show, when z = 1, the value of y when x approaches infinity is e. And when z=2, the value of y when x approaches infinity is 1. So, what is the limit for y for values of z between 1 and 2? The interesting answer is that y=e only at x=1. For x1, y=1. If you graph this, you get a right angle, with the vertix at (1,1).

are you real?

I like viewing things through complex/harmonic analysis, exponents are scaling/rotations, and exponents of 1 are just the null rotation around the origin. An infinite sequence of null rotations is just θ + 0 + 0 + 0 ... = θ

Yes in calculus we deal with such problems where f(x) and g(x) are two functions such that f(x) ^ g(x) such that f(x) tends to 1 and g(x) approaches to infinity when 'x' tends to infinity........ so here we can use the limit concepts such that limit 'x' approaches to infinity where f(x) tends to 1 and g(x) tends to infinity the we take " e^ lim x approaches to infinity ln (g(x). (f(x)-1) ) " .....and by applying limit we could find answer

(1+a)^N with a>0 goes to infinity, for -1

idea: what about the 0 root of a number? I saw that thi is a number elevated at the 1/0 power, which tends to +/- infinity, but the calculator says 0. Could you explain it?

I once read a sybolic logic book that took 137 pages to come to : "therefore we can assume that 1+1=2" So anything is possible if you put enough mental gymnastics into it.

Indeterminate forms in mathematics are a treat to watch damn

Love your videos

It's interesting that my calculator calculates 1^♾️ as 1, but puts a note that "1^♾️ is replaced by 1"

There is no exponent Infinity since the exponent has to be a number. However Infinity is not a number.

You can also look at the limit as x approaches 1 of x^infinity. For any number greater than 1, the result is infinity. For any number less than 1, the result is 1/infinity or 0. So 1^infinity is a discontinuity in that plot. 1^infinity is indeterminate. But there are contexts where one type of evaluation is appropriate, and in those situations, you can use a fixed value.

1:35 is completely wrong. Those terms are not 0, they are slightly bigger than 0 and thus you have a slightly bigger number than 1 over infinity that's why it does not work. If you have a plain exact 1 over infinity, it's 1. You can't use normal calculations for special cases like that and then say that the definition is wrong. Your calculations are wrong.

can you do a video about multivariable calculus and it's applications and cool properties?

Infinity is the most interesting number even though it's not a number

I am not 100% certain I know how this distinction works. So please correct me if I'm wrong in any assumptions or conclusions I make below: Is there a scenario where the expression 1 ^ ∞ does not imply limits? Since obviously infinity is not a number. In that case wouldn't there be just 2 possible definitions: 1) 1 ^ ∞ = lim(x->inf) 1 ^ x or 2) 1 ^ ∞ = lim(x->inf)lim(y->1) y ^ x ? As you mentioned in your example of the definition, the limit in 1) is just 1, while the limit in 2) is indeterminate. But isn't the first definition much more intuitonal? Of course, when you use infinity arythmetically, you imply a limit related to that infinity., But it's (in my view) totally counterintuitive to suspect 1 of being actually a limit of something approaching 1. If we made assumptions that way, we would have to suspect every number that is seemingly constant as not being the number itself, but a limit to that number. This means implying we're taking a limit for every unique number in the expression [which might look like 2) but with more limits depending on the expression] While I do understand the meaning and importance of the very prevalent "1 ^ ∞ indeterminate form", I do think that 1 ^ ∞ alone implies a constant 1, not a limit to 1, because it would need to be specified as such, so 1 ^ ∞ = 1.

So, 1^infinity tends to 1, but let's take for example (1 + 1/x)^x as x tends to infinity. Because of calculus we know that this tends to e (in fact the exact definition of e is this limit here), but if we try to solve this limit it it turns into 1^infinity. But It JUST __TENDS__ to 1^infinity without actually being exactly 1^infinity, and that's why the definition cannot work, right? Because the EXACT 1^infinity tends to 1, but anything else that just TENDS to 1^infinity can just tend to anything else and not to 1^infinity which is 1, right?

My math training is limited (and decades-old), but a number to the power of infinity being undefined strikes me as a matter of convention, not fundamental mathematical "truth" (assuming the latter concept actualy makes sense...) I can't know for sure, but would a conventional definition that defined it to be equal to the limit of the number to the power of x as x grows without bound ACTUALLY result in problematic mathematical outcomes?

It also depends what u mean by the symbol infinity. Is it some aleph cardinality or for example compactyfying point in topology. In cardinal arithmetic in set theory u can well define exponents with cardinals.

So what if I do something like this. I choose an abitrarily large number x. I define y as floor(x). z as x - floor(x). Then I define f(x)=1^x. If x is positive, then f(x)=1*f(x-1). I can do something similar for negative values. Then by induction, f(x) is 1, right? 1^(x+y)=1^x*1^y I think. And for all y > -1, and < 1, 1^y is also 1. Weak azz proof I know, but where am wrong on this?

I was thinking that for sooo many days thx for vedeo

(Note: inf means infinity) There is a person saying you all are wrong but, to be honest, the core of the problem just lies in the different ways to interpret the expression '1^inf' (1) Most of us regard it as lim n->inf (1^n), which should be 1. (From the RU-vidr at 3:50) I can (TRY TO) prove it via the rigorous definition of limit. Let c > 0 be arbitrary. For every c, there exists a natural number N (Trivially choose N = 1 for any c) such that n >= N implies |1^n - 1| = |1 - 1| (n is a natural number) = 0 < c Equivalently, lim n->inf (1^n) = 1. (2) However, as the RU-vidr (at 0:55) unveils in his video, '1^inf' means lim f->1,g->inf (f^g), which indeed has different values depending what f and g are, so it is indeterminate. (3) Lastly, as a commenter insists (and somewhat mentioned by the RU-vidr at 1:57), the '1^inf' literally means '1^inf'. As I spent like 1 and a half hour reading the comments, that 'inf' is probably the 'inf' in the 'extended real number set'. 'inf' is the supremum of the 'real number set' R. It is an object but doesn't have nice arithmetic properties so that we can 'find' a value for '1^(inf)', i.e. it is undefined. Well. I probably get it since we should have 'real number set' R := (-inf,inf) and the 'n -> inf' in (1) probably lies in R and we can just preceive n as some real number and trivially say '1^(inf) = 1' Yet, with 'inf' actually being 'inf' as mentioned in (3), the arithmetics are totally different since we are not talking about R =: (-inf,inf) anymore, thus '1^(inf)' is not defined. (Not in accordance with the rules in R especially) So, I, as a high schooler, reckon the problem lies in how we interpret '1' (causes confusion between (1) and (2)) and 'inf' (causes confusion between (1) and (3)) in '1^inf'.

How my math teacher described how e was defined: "If you mutiply enough 1s together, eventually you get 3"

The best way i can describe it is 1^inf is always 1 if the base is 1 however in the case like the limit definition of e which is limit as x approaches inf of (1+1/x)^x the base 1+1/x doesnt equal 1 it approaches 1+ its approaching 1 but while it approaches 1 its always a number larger than 1 thats why its indeterminate you are taking a function that approaches 1 and raising it to a function that approaches infinity

clearly the answer is -1/12

I am not a mathematician but i cant believe that if we do 1*1 even until the end of time, or what ever infinity means, will somehow the result be changed. If you can tell me a reason, i am all ears..

Very intetesting you said you'd done 0^* Have you done one on infinity^0 or is it an easy one?

It is somewhat defined to be 1 by rule that every value can be multiplied by one and not change, in the e example we have limit that aproaches 1 but it is not equal 1. I mean in some region of math infinity just should mean that something never ends for example anything divided by 0 excluding itself should be infinity becouse we can always substract 0 forever and not end up with 0. I know It would mess up with math in general but it should have its own category

Yeah my teacher showed this to us in school. Cool

Fun fact: if you put x^infinity+y^infinity=1 in desmos, it draws nothing, but if we put x^infinity+y^infinity

One, who calculates lim (1+1/x)^x = Lim (1)^x = 1 for x ->oo shouldn't pass Calculus 1. The comparison of (Const 1)^oo to (1+f(x))^x with f(x) -> 0 for x-> oo is misleading, since all x's have the same value at the same time, u can't expect Lim (1 + Lim f(y))^x to be the same...

Interesting approach to define 1∞ as indefinite is looking at the formula a∞. For all a∞(a^x)=0, for all a>1, lim x>∞(a^x)=∞, so there's a discontinuation at a=1 similar to 1/x for x=0.

It will be either one or undefined I guess

Calculator: *Domain Error* The answer to the problem I put in:

Nice! Thanks for sharing friend!

I think the limit as x->∞ (1^x)=1 But if you have anything in the function that approaches 0 you need to remember it’s 0^+ or 0^-. Meaning an arbitrarily small value, arbitrarily close to 0, making the limit indeterminate.

ω is the variable that represents ordinal infinity. Ordinal just means there's a number before, ω-1, and a number after, ω+1. ω and others like it belong to a class of numbers called transfinite. You can do math to ω; ω+2 does not equal ω, instead it is two more than ω. There are countably infinite numbers between 0 and ω. In fact, there are precisely ω-1 whole numbers between 0 and ω. While ∞-∞ = NaN, ω-ω = 0. I'd imagine that 1^ω would indeed be 1, instead of NaN. What is it with 0, ∞, and their copious undef and NaN results? Is math broken? Was something big overlooked? Is there another imaginary plane of numbers where 1/0 is another i?

Try to put an infinity symbol in any purely arithmetic expression and you're bound to mess up. As said in the video, infinity is not a number. So asking such questions, at least to me, feels weird anyways. I don't expect any arithmetic to work if I substitute a number with something "not-numberly". Like, what's one raised to the apple? Tho, I'm surprised not to see an argument using the power series in this video. That would get out of hand decently quickly having to deal with ∞^(∞×0). (Which again sounds to me like "Apple raised to the power of apple but for the case that you never had apples anyways") Thanks for the video, keep up the good work ^~^

Great idea for a math discussion!

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This guy is AI generated

quick info: this guy really hate undefined things

i think theres a problem saying lim(x->inf) (1+1/x)^x = 1^inf the power law of limits says you can distribute the limit into the parenthesis and assume the lim of the inside and take the power of that. But that's for fixed powers. I don't think you are allowed to assume here that the lim can go inside, take that limit, and then also take the limit of the power. And so I don't think you can assume that lim(x->inf) (1+1/x)^x is the same as (lim(x->inf)(1+1/x))^x =? (1)^x . The answer is ofc e, which you get by doing that limit properly maybe there's other ways to argue why it's indeterminate, but i disagree with this one as for the log, that was just log(1^x) and plugging in x=inf, which is also not the proper way to do limits

One of the fun things that you can do in math is just axiomatically state x^inf=1 as a rule.

You can also approach it with the limit of 1^x, with x-->inf. Oh. You just did the straightforward thing after taking the totally BS path. Well done.

Indeterminate

Correct me if I'm wrong but if you find the derivative of 1^x you get 1^x * ln(1) which equals 1^x * 0 which equals 0. This means that the function 1^x has a rate of change of 0 so if 1^1 =1 then for all values of x it will be 1

To accurately determine the value of 1^∞, you would need more information about the specific function or expression in which it appears. It's crucial to consider the context and apply appropriate techniques, such as L'Hôpital's Rule, limit laws, or other methods from calculus, to evaluate the limit.

it feels to me kinda similar to 0^0. It is an indeterminate form, but we "define" it to be the limit as x goes to 0 (from above) of x^x, which is 1

1 elevated to inf is 1, the problem was not saying something close to 1, but exactly 1, those other limits are a whole different problem which is the case when something approaches to 1

related question: does the proof by induction work up to infinity? if so, it’s easy to show that the result is 1. if not ( for example because inf. is not a natural number), why the proof by induction is usually used to prove things for infinite cases?

But Lim( 1^x ) as x gets large is 1, because the sequence 1, 1^2, 1^3, 1^4, ... is a cauchy (and therefore convergent sequence) such that the limit is bounded between 1 and 1, so it tends to 1.

What is the zeroth root of 1 please help

I believe using infinity as an exponent is like saying one to the power of an apple, a random object from real life used in mathematical calculation as not a number

I want to ask what's won't with taking infinity as 1/0. I think it works in every way. It we do the same for this problem, we get 1^infinity is any real no. because x^0=1.

Sorry but this is wrong. 1 to the power infinity should only be considered indeterminate when we have f(x)^x where x goes to infinity and f(x) goes to 1, and f(x) is not equal to 1. In that case the issue is that we don't know if f(x) goes to 1 "fast enough" to cancel out the effect of the exponent, therefore the result is indeterminate. But if f(x) = 1, then we can prove by complete induction that the result is one.

We know its 1 White out any calculation your title of video is 100% wrong bc i expect it would be 1 and in end of video you show is 1

This video is mathematical misinformation, and I ask that you take it down or add a significant disclaimer to the description. Around the one minute mark, you begin discussing how 1^inf can be described using limits, but don't actually look at the limit of 1^x as x --> inf. Instead, you swap in (1 + 1/x)^x and (1 + 1/x^2)^x in a bit of flimflam. The misinformation is most apparent at the 1:30 mark, in the sentence beginning, "Essentially..." The statement you make there is simply *false*; you use flawed logic and violate the definition of a limit and the rules we follow to evaluate them at infinity, and then use this to support why 1^inf is indeterminate. Please be clear: the limit of 1^x as x --> inf is 1, period. For any sequence of x that diverges to infinity, the sequence 1^x converges to 1; in fact, every element of the sequence is 1. It might be that 1^inf is considered indeterminate in mathematics for *other* reasons, but the interpretation of this statement as a matter of limits is clear. Please don't put up pseudo-educational videos like this.

@1:46 but we know the harmonic series diverges, so that doesn't "look like" 1^inf to me. And "Looks like" doesn't sound very mathematically rigorous to me. To me, what you did is demonstrate that (1+1/inf)^inf is NOT 1^inf.

So effectively 1^inf = 1 in the same way i = sqrt(-1). It isn't really but everyone understands what you mean and can work with it (though I would love to see a single example of anyone ever finding a use for 1^inf xD). Though now comes the real question. Since I've been taught 3*0.3333333 is defined as 1, is that the something that breaks 1^inf? I'd imagine 0.9999999 recurring to the power of infinity would approach zero because it's never quite there yet, but since it is defined as actually fully 1 it would be expected to stay 1 at the same time :p I guess you did a similar example though except this would be (1-1/x)^x for x-> infinity and it'd converge towards 1/e according to wolfram alpha.

1 raise to infinity is equal 1.this is simple math.

Why use the functions (1+1/x)^x and (1+1/x^2)^x in the limit analysis, when the more obvious function 1^x is not used? Not to say evaluating part of the function to its limit before evaluating the whole function’s limit. Seems just an underhanded way to dismiss the approach of using limit analysis to find 1^infinity.

Seems like an example of overthinking this, to infinity.

i feel like it makes sense because no matter how many times you multiply 1 by 1, its not going to get any bigger. the same way i know 2 multiplied by 2 infinitely will never have a prime factorial of 3 or 5.

A more interesting definition from set-theory is that a^b := |b -> a| where you let the numbers a and b be sets of a and b elements respectively in the second half. That is, the power-function is defined by the total number of functions between sets of a given size. Or reversed, the number of functions between two sets A and B, |A -> B| = |B|^|A|. It gives you the normal values like: x^0 = 1, 0^y = 0, if y > 0 and everything else when base and exponent are natural numbers. But for the case of 1^infinity, you can use that (N -> {1}) only contains the constant function f(n) := 1 to say that 1^infinity = 1.

1^∞ is 1 for most values of 1 used as input. For values that only approach 1 it can be any non-negative finite value.

1 to the power of infinity is only indeterminate if 1 is the approach, if you only apply the variable in the exponent it is 1 and even the log of it is only indeterminate when you use limits on the obviously static parts rather than infinity which is the only approaching number

I have yet to find a video that can explain this to me in a way I understand. The best I found was assuming people are not talking about the integer 1. But that's annoying because why wouldn't I assume that you are talking about the integer 1. If you were talking about ~1 then yes this video would make sense because below or above 1 leads to an impossible graph as x approaches infinity, but if I am talking about the integer 1, no floats, no guess work, the whole value 1, then yes 1 is 1 to the power of infinity. But it seems like at some point you need to view 1 as function and not an integer and there seems to be no consensus there. Unless I missed the lesson that when infinity is included you are not allowed to assume any number is whole or an integer just a function that approaches that number.