# 1 to the power of infinity, why is it indeterminate?

• B
Gold Member

## Main Question or Discussion Point

I've been taught that $$1^\infty$$ is undetermined case. Why is it so? Isn't $$1*1*1...=1$$ whatever times you would multiply it? So if you take a limit, say $$\lim_{n\to\infty} 1^n$$, doesn't it converge to 1? So why would the limit not exist?

PeroK
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I've been taught that $$1^\infty$$ is undetermined case. Why is it so? Isn't $$1*1*1...=1$$ whatever times you would multiply it? So if you take a limit, say $$\lim_{n\to\infty} 1^n$$, doesn't it converge to 1? So why would the limit not exist?
The indeterminate case is ##\lim_{n \rightarrow \infty} a_n^{b_n}## where ##\lim_{n \rightarrow \infty} a_n =1## and ##\lim_{n \rightarrow \infty} b_n =\infty##

##\lim_{n\to\infty} 1^n = 1##

scottdave, Delta2 and fresh_42
fresh_42
Mentor
I can only guess, what the one who said "undetermined" has meant.

##\lim_{n \to \infty} 1^n = 1## is certainly true. But this doesn't allow us to write ##1^\infty##. What does it mean? It is just not defined and as such, undetermined. You read it as ##1^{\mathbb{N}}##, but why isn't it ##1^{\mathbb{R}}##? Fact is, we can only finitely often multiply ##1## by itself, so ##1^\infty## is not defined.

dextercioby
Gold Member
I can only guess, what the one who said "undetermined" has meant.

##\lim_{n \to \infty} 1^n = 1## is certainly true. But this doesn't allow us to write ##1^\infty##. What does it mean? It is just not defined and as such, undetermined. You read it as ##1^{\mathbb{N}}##, but why isn't it ##1^{\mathbb{R}}##? Fact is, we can only finitely often multiply ##1## by itself, so ##1^\infty## is not defined.
But doesn't 1^infty mean that we are multiplying 1 infinite times and hence the final product must be 1.

TachyonLord and Delta2
PeroK
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Gold Member
But doesn't 1^infty mean that we are multiplying 1 infinite times and hence the final product must be 1.
No. ##\infty## is not a number, so ##1^{\infty}## is simply not defined. You can't do anything an infinite numbers of times, which is why limits were developed.

TachyonLord, WWGD, QuantumQuest and 2 others
Delta2
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Gold Member
it would be useful to read again what @PeroK wrote at post #2, it is important to distinguish between the two cases

1) ##1^{\infty}## which is we could write it more formally as ##\lim_{n\to \infty}1^n## and which equals 1

2) ##\lim_{n\to \infty} a_n^{b_n}## with ##\lim a_n=1## and ##\lim b_n=\infty##. This case is different than the case 1) and it is indeterminate. Analogous is the difference where we have ##0\cdot\infty## . It is ##\lim 0\cdot n=0## but also ##\lim \frac{1}{n}##=0 ,##\lim n=\infty## but ##\lim \frac{1}{n}\cdot n=1##.

if we take the logarithm of ##a_n^{b_n}## we will have ##b_n\ln{a_n}## and it would be
##\lim b_n\ln{a_n}=\lim b_n\lim\ln{a_n}=\infty\cdot \ln1=\infty\cdot 0## that is it reduces to the case ##0\cdot\infty##

fresh_42
Mentor
##1^\infty## doesn't tell you what kind of infinity it is, and what to do with the number ##1##. You can write
$$\prod_{n=1}^\infty 1^n = 1$$
but this is an abbreviation of ##\lim_{n \to \infty}1^n = 1##
... which is why limits were developed.
I think one could still agree on ##1^\infty = 1## if everybody agrees on the notation ##1^\infty = \lim_{n \to \infty}1^n## but then in my opinion it will do more harm than good, because of the example @PeroK has given: You cannot substitute limits of sequences in arithmetic expressions if they are not numbers, or the expression isn't defined. ##1^\infty## would suggest that ##\lim_{n \to \infty}a_n^{b_n} = 1## if ##a_n \to 1## and ##b_n \to \infty## which is wrong.

I think this example as in post #2 is the reason why ##1^\infty## should not be used.

Mark44
Mentor
I've been taught that $$1^\infty$$ is undetermined case. Why is it so?
Perhaps you meant the indeterminate form ##[1^\infty]## (written in brackets to emphasize that this is an indeterminate form).
This limit -- ##\lim_{n \to \infty}(1 + \frac 1 n)^n## -- is an example of this indeterminate form. Naively, you might think that since the quantity in parentheses is approaching 1 as the exponent increases without bound, the value of the limit is just 1. This isn't true -- the actual limit is the natural number ##e##.

scottdave and suremarc
lurflurf
Homework Helper
I can only guess, what the one who said "undetermined" has meant.
If you were to guess it would be reasonable to guess the one who said "undetermined" has meant what calculus books mean when they use "undetermined" as it is standard terminology.
$$1^\infty$$
means
$$\lim a^b$$
where
$$\lim a=1$$
and
$$\lim b=\infty$$
You cannot substitute limits of sequences in arithmetic expressions if they are not numbers, or the expression isn't defined.
You have not defined numbers, nor is it a term with a widely agreed upon precise meaning.
Arithmetic expressions are not real numbers would be a more sensible statement.
We can define an arithmetic of limits of sequences.
We can compare this system to real number arithmetic or extended real number arithmetic.
For example by writing
a={a}
b={b}
where a and b are limits of {a} and {b} respectively.
Often
$$\{a\}^{\{b\}}=\{a^b\}=a^b$$
ie
if
{a}=3
{b}=5
$$\{a\}^{\{b\}}=\{a^b\}=3^5=243$$
but not always.
Indeterminate forms are the situations where this does not hold and more detailed analysis is needed.
$$1^\infty$$
is one example.
$$\{(1+1/n)^n\}=e\ne \{(1-1/n)^n\}=1/e$$
are unequal even though
$$\{1+1/n\}=1= \{1-1/n\}$$
To give one example.

What does $1\cdot 1\cdot \ldots$ mean exactly? By definition, the binary operation ##\cdot## is guaranteed to 'work' for finite expressions i.e you can apply it finitely many times and you have something that's well-defined. That's not the case for multiplication over arbitrary families so you have no guarantee that, for example ##\prod _{n\in\mathbb N} 1 ## or ##\prod _{x\in\mathbb R} 1 ## are well-defined.

For all positive x in ℝ, $\lim\limits_{n\to\infty} x^\frac{1}{n} = 1$.

WWGD
Gold Member
2019 Award
You can , in a sense, substitute:

##1^{a_n} = e^{a_nln(1)} ## and then try L'Hopital : ##e^{0/(1/a_n)}##. You need a few details worked out.

The indeterminate case is ##\lim_{n \rightarrow \infty} a_n^{b_n}## where ##\lim_{n \rightarrow \infty} a_n =1## and ##\lim_{n \rightarrow \infty} b_n =\infty##

##\lim_{n\to\infty} 1^n = 1##

That's fascinating!! I first encountered the statement "0x∞ is indeterminate" and never thought twice of that; but actually this is not indeterminate, right?

limn->∞ 0.n = 0

while the real indeterminate expression is this, right?

limn->∞ limm->0 m.n

Learning something new in my old years, yeah!

Hmmm... something not making sense in what I just posted. Specifically... shouldn't we write that n and m go to their limits simultaneously? Like, something like this, maybe (although I never saw this notation, I'm inventing):

limn->∞, m->0 m.n

Or is the "lim" thingie something that we can compound, just like a function? like, is this true?

limn->∞ limm->0 m.n = limn->∞ ( limm->0 m.n ) =
limm->0 ( limn->∞ m.n ) = limm->0 limn->∞ m.n

and all of that is the same as

limn->∞, m->0 m.n ?

suspect that expressions like the one below don't make sense, because it implies the variables are going to their limits separately (it assumes independent variables); there's no such thing as the "limit of two independent variables", right?

limn->∞ limm->0 f(n).g(m)

Therefore I feel I've butchered Mathematics with that notation; I guess the right notation (repeating DeltaK above) is this form:

limn->∞ f(n).g(n)

where limn->∞ f(n) = 0, limn->∞ g(n) = ∞

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WWGD
Gold Member
2019 Award
That's fascinating!! I first encountered the statement "0x∞ is indeterminate" and never thought twice of that; but actually this is not indeterminate, right?

limn->∞ 0.n = 0

while the real indeterminate expression is this, right?

limn->∞ limm->0 m.n

Learning something new in my old years, yeah!
limn->∞ 0.n = 0 this is true only of/when the limit exists, i.e., it is finite.

You can , in a sense, substitute:

##1^{a_n} = e^{a_nln(1)} ## and then try L'Hopital : ##e^{0/(1/a_n)}##. You need a few details worked out.
Isn't the following the actual reason? It's basically the exponential equivalent of dividing by 0 (taking something to the 1/infinitieth power):
For all positive x in ℝ, $\lim\limits_{n\to\infty} x^\frac{1}{n} = 1$.

jbriggs444
Homework Helper
2019 Award
Hmmm... something not making sense in what I just posted. Specifically... shouldn't we write that n and m go to their limits simultaneously? Like, something like this, maybe (although I never saw this notation, I'm inventing):

limn->∞, m->0 m.n
That is a reasonable notation. But now you have to worry about the path which is traversed as n and m each go toward their respective limiting values. If the result can depend on the path then the limit is indeterminate.
limn->∞ limm->0 m.n = limn->∞ ( limm->0 m.n ) =
limm->0 ( limn->∞ m.n ) = limm->0 limn->∞ m.n
Limits do not, in general, commute.

Matt Benesi
WWGD
Gold Member
2019 Award
Isn't the following the actual reason? It's basically the exponential equivalent of dividing by 0 (taking something to the 1/infinitieth power):
Yes, this is a good equivalence. But notice that we know the denominator of the exponent will not, however small, ever by zero, so that the exponent is of the form 0/b with b non-zero. And then we can use continuity of exp.

Yes, this is a good equivalence. But notice that we know the denominator of the exponent will not, however small, ever by zero, so that the exponent is of the form 0/b with b non-zero. And then we can use continuity of exp.
Ok, I'm not following what you're saying. Can you explain it more?

I thought in the case of ℝ, that $1^\infty$ was undefined because for all positive x in ℝ $\lim\limits_{n\to\infty} \, x^\frac{1}{n} \, = \, 1$.

So $\lim\limits_{n\to\infty} \, 2^\frac{1}{n} \, = \,\lim\limits_{n\to\infty} \, 3^\frac{1}{n} \, = \,\lim\limits_{n\to\infty} \, 4^\frac{1}{n} \, = \, b$ so $\lim\limits_{n\to\infty} \, b^n \, = \, 2 \, = \, 3 \, = \, 4$, etc. unless limits are supposed to be specific, so $\lim\limits_{n\to\infty} \, 2^\frac{1}{n} \, != \,\lim\limits_{n\to\infty} \, 3^\frac{1}{n} \,$.

What about ℂ? $1 \, = \, ( i^x )^\frac{4}{x} \, = \,e^{i \, \frac{\pi}{2} \cdot x \cdot \frac{4}{x}}$, for all x in ℝ.... but whats up with the 0 case?

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WWGD
Gold Member
2019 Award
Ok, I'm not following what you're saying. Can you explain it more?

I thought in the case of ℝ, that $\lim\limits_{n\to\infty}1^n$ was undefined because for all positive x in ℝ $\lim\limits_{n\to\infty} \, x^\frac{1}{n} \, = \, 1$.

So $\lim\limits_{n\to\infty} \, 2^\frac{1}{n} \, = \,\lim\limits_{n\to\infty} \, 3^\frac{1}{n} \, = \,\lim\limits_{n\to\infty} \, 4^\frac{1}{n} \, = \, b$ so $\lim\limits_{n\to\infty} \, b^n \, = \, 2 \, = \, 3 \, = \, 4$, etc. unless limits are supposed to be specific, so $\lim\limits_{n\to\infty} \, 2^\frac{1}{n} \, != \,\lim\limits_{n\to\infty} \, 3^\frac{1}{n} \,$.

What about ℂ? $1 \, = \, ( i^x )^\frac{4}{x} \, = \,e^{i \, \frac{\pi}{2} \cdot x \cdot \frac{4}{x}}$, for all x in ℝ.... but whats up with the 0 case?
Sorry if I was confusing, Matt, but we may have to spend some time unwrapping this: are you referring to EDIT ## Lim _{ a_n \rightarrow \infty} 1^{a_n} ## or just the expression ## Lim _{ n \rightarrow \infty} 1^n ##?

Sorry if I was confusing, Matt, but we may have to spend some time unwrapping this: are you referring to ## Lim _{ n \rightarrow \infty} 1^n ## or just the expression ## Lim _{ n \rightarrow \infty} 1^n ##?
Ohhh. I get what you said earlier now (the cases I mentioned weren't specific enough):
You can , in a sense, substitute:

##1^{a_n} = e^{a_nln(1)} ## and then try L'Hopital : ##e^{0/(1/a_n)}##. You need a few details worked out.

WWGD