Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question already has an answer here:

$1^\infty$ = undefined

$2^\infty = \infty$

$0^\infty = 0$

Why is $1^\infty$ undefined? People were trying to explain to me that infinity isnt part of the Real numbers, yet, $2^\infty$ and $0^\infty$ somehow ARE defined?

In my opinion $1^\infty = 1$. I mean isn't it not easy to prove since $1\times 1\times 1 \times \cdots=1$ no matter how many times you do it?

share|improve this question

marked as duplicate by Andres Caicedo, Ross Millikan, Grigory M, Austin Mohr, egreg May 7 at 22:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1  
Who told you that $2^{\infty}$ is well defined (in $\mathbb{R}$)? –  nsanger May 7 at 21:33
1  
this is outputs from wolframalpha.com –  SN1 May 7 at 21:34
1  
To say that something "equals undefined" is at best very very clumsy language. One should say that an expression is undefined. This is the "'is' of predication", not the "'is' of equality". –  Michael Hardy May 7 at 21:43
add comment

8 Answers 8

It's best not to think of $1^\infty$ as a value, but rather as a shorthand - the same can generally be said anywhere that you see $\infty$ in this sort of analytical context. For instance, $0\cdot\infty$ (usually) isn't the actual product of these two entities, but rather a shorthand for an expression of the form $\lim\limits_{n\to\infty}(a_n\cdot b_n)$, where $\lim\limits_{n\to\infty}a_n=0$ and $\lim\limits_{n\to\infty}b_n\to\infty$ (that is, $b_n$ grows unboundedly). Similarly, $\dfrac00$ usually refers to an expression of the form $\lim\limits_{n\to\infty}\left(\dfrac{c_n}{d_n}\right)$ with $\lim\limits_{n\to\infty}c_n=0$ and $\lim\limits_{n\to\infty}d_n=0$; etc.

Part of the reason that the shorthand works is that it preserves the maning of any well-defined expression (in fact, that can be used as a definition of well-defined); we can equally well say that $1\cdot 2=2$ 'in shorthand' because if $\lim\limits_{n\to\infty}a_n=1$ and $\lim\limits_{n\to\infty}b_n=2$ then $\lim\limits_{n\to\infty}(a_n\cdot b_n)=2$ also. When people say that $1^\infty$ isn't well-defined, what they're really saying is that this shorthand falls down and doesn't give us a unique value.

share|improve this answer
2  
Arguably, $\pi$ or $\sqrt{2}$ or even $3/4$ is just as much of a "shorthand" as $1^{\infty}$ is. Preventing yourself from forming a concept of arithmetic with these things by insisting it's all really shorthand for something more complicated will just make things more difficult in the long run. –  Hurkyl May 7 at 21:51
    
@Hurkyl It's certainly true that those can all be considered shorthands, but they're (at least to me) shorthands in different ways; I generally think of something like $\sqrt{2}$ as a shorthand for a definition of a thing, whereas $1^\infty$ is a shorthand for a (limiting) process; it's in some fundamental sense dynamic rather than static. –  Steven Stadnicki May 7 at 21:55
    
@StevenStadnicki: It all depends on how you look at number systems. If you accept the existence of rational numbers only, then $\sqrt{2}$ is a shorthand for a limiting process too (I'm referring to completion of $\mathbb{Q}$ by Cauchy sequences). –  math.n00b May 7 at 21:58
1  
@Steven: In my opinion, one of the biggest conceptual difficulties people have with calculus is that they can't think of a limit as a "thing". e.g. you see all sorts of people having problems with $\lim_{x \to 0} x = 0$ because "$x$ doesn't actually reach zero: it's merely getting closer to $0$". I believe viewing $2^\infty$ as a limiting process rather than a thing is really the same problem: it's just obscured because it's easy to shift the blame to the fact the "thing" isn't an ordinary real number. –  Hurkyl May 7 at 21:59
add comment

These rules encode information about limits of sequences. If you are trying to evaluate $\lim_{n \to \infty}(a_n)^{b_n}$ and are told $a_n \to 0, b_n \to \infty$ you can be sure the limit is zero. Similarly, if you are told $a_n \to 2, b_n \to \infty$, you can be sure the limit is $\infty$. If you are told $a_n \to 1, b_n \to \infty$, you don't have enough information. In that way it is similar to $\lim_{n \to \infty}\frac{a_n}{b_n}$ where $a_n,b_n \to 0$. You need more information to evaluate the limit. It could be $0$, it could be some finite number, it could be $\infty$

share|improve this answer
add comment

$1*x=x$ for any $x \in \mathbb{R}$, but that is true only if the "$1$" you're talking about is exact, not in a limit sense.

Consider this famous limit $$\lim_{n \to \infty} (1+\frac{1}{n})^n$$

This is $1^{\infty}$, but it's limit is a very important mathematical constant denoted by $e$.

You can convince yourself that $e \neq 1$ by calculating that limit using a calculator. Try to plug in positive integers for $n$ in $\displaystyle A_n= (1+\frac{1}{n})^n$ and you'll see that the value of $A_n$ is getting bigger and gets closer and closer to $2.718\cdots$

share|improve this answer
    
yea people got mad at me for pointing that out math.stackexchange.com/questions/785144/… –  SN1 May 7 at 21:44
    
@SN1: Well, This might seem bizarre at first, because you think that $1^{\infty}$ is equal to $1$. But the truth is, that $\infty$ is not a number, and whatever that deals with $\infty$ needs to be analyzed carefully. If you want to get convinced, just plug in numerical values instead of $n$ and find the limit by a calculator. –  math.n00b May 7 at 21:47
add comment

Because $+\infty$ is an element of the extended real numbers. That number system is extremely useful in calculus and analysis, which is why people explain things using them.

Continuity is a key idea in these fields, and we often prefer to continuously extend functions whenever possible. e.g. we often like to continuously extend the function $f(x) = x/x$ to have the value $f(0) = 1$. And when we use the extended real numbers, we like to continuously extend all sorts of functions to have values at $+\infty$ and $-\infty$ when appropriate.

For any $x > 1$, you have $x^{+\infty} = +\infty$. This is right because, for example,

$$ x^{+\infty} = \lim_{\substack{y \to x \\ n \to +\infty}} y^n $$

And for $0 < x < 1$, you have $x^{+\infty} = 0$ for the same reason. (this post will not comment on defining $0^x$)

But because $1$ lies on the boundary between where the value of $x^{+\infty}$ should be $0$ and where it should be $+\infty$, you can't continuously extend the function $x^n$ to $x = 1$ and $n = +\infty$. Thus, we leave it undefined.


As an aside, if you were thinking specifically of $f(x) = 1^x$ rather than $g(x,y) = x^y$, then it would make sense to define $f(+\infty) = 1$, even though we need to leave $g(1, +\infty)$ undefined.

share|improve this answer
    
By that same logic e.g. $x^2 / x$ at 0 is also $0/0$; but it should be 0, not 1. –  vonbrand May 7 at 22:02
    
Yes, lim x->0 of x^2/x is equal to zero, so if we were continuously extending f(x) = x^2/x, we would say the new function f'(0) = 0. –  Zach Effman May 7 at 22:15
    
@vonbrand: Right: that's why we don't extend $f(x,y) = x/y$ to have a value at $f(0,0)$. –  Hurkyl May 7 at 22:33
add comment

I woul have thought if anything $1^\infty$ was defined as anything to the power of $1$ is $1$? The same for $0^\infty$ as $0$ to the power of anything is still $0$. However $2^\infty$ would be undefined as the real numbers are an open set so it just gets infinitely larger as your power tends towards $\infty$?

share|improve this answer
add comment

You can think in the limits of $0^n$, $1^n$ when $n$ tends to infinity. And with $2^n$ you can think in a unbounded positive sequence.

share|improve this answer
add comment

Consider $u_\epsilon \to a>1$ and $v_\epsilon\to b<1$, and $A_\epsilon\to\infty$.

Then $u_\epsilon^{A_\epsilon} = \exp [A_\epsilon \log u_\epsilon] $;

$\log u_\epsilon\to\log a>0 $ then $$ A_\epsilon \log u_\epsilon\to\infty \\ u_\epsilon^{A_\epsilon}\to\infty $$

If you replace $u_\epsilon$ with $v_\epsilon$ you get

$$ A_\epsilon \log v_\epsilon\to -\infty \\ v_\epsilon^{A_\epsilon}\to 0 $$

share|improve this answer
add comment

Since it seems like no one else has emphasized this:

$1^{\infty}$ is not well defined because $\infty$ is not a number.

When we do $2^3$ we are multiplying $2$ by itself $3$ times and we get the real number $8$. We extend the definition of $a^b$ so that it works for any positive real number $a$ and any real number $b$. And $\infty$ is not a real number. So $a^\infty$ is not well defined no matter what $a$ is.

It is the same with $\infty + 2$. Again this is not defined because $\infty$ is not a number.

Likewise $2^{\infty}$ is not well defined.

Here is the catch though: We can make sense of things like $\infty + 2$ by talking about the extended real number line. We can also make sense of $\infty$ when we talk about limits.

So for example $$ \lim_{x\to 0^+} \frac{1}{x} = \infty. $$ Things like this have precise definitions. And it is tempting to write the above as $$ \frac{1}{0}. $$ But this is not defined. We are only allowed to divide by non-negative real numbers.

That said, you can find various online sites that will say that $\frac{1}{\infty} = 0$. What they mean is that $$ \lim_{x\to \infty} \frac{1}{x} = 0. $$ This means that you can't necessarily trust these online sites when you are trying to carefully write down a mathematical argument.


Now to try to get at what is going on with saying that $1^{\infty}$ it is also worth pointing (as is done in other answers) that we can interpret $1^{\infty}$ differently in the context of limits.

For example consider: $$ \lim_{x\to 1^+} x^{\frac{1}{1-x}}. $$ Here the base approaches $1$ and the exponent approaches $\infty$. If we want to actually calculate this limit, we have to work hard than just appealing to some general definition of $1^{\infty}$. We can solve the problem using logarithms: $$ y = x^{\frac{1}{1-x}}\\ \ln(y) = \ln(x^{\frac{1}{1-x}}) = \frac{\ln(x)}{1-x} $$ and then find (using L'Hopital's rule) $$ \lim_{x\to 1^+} \ln(y) = \lim_{x\to 1^+}\frac{\frac{1}{x}}{-1} = -1. $$ So $$ \lim_{x\to 1^{+}} y = e^{\ln(y)} = e^{-1}. $$ Likewise it is possible to find similar examples where the limit is any given real number. You can even make the limit infinity.

share|improve this answer
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.