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

In summary, the limit of a sequence does not always exist and can only be defined for certain finite sequences.
  • #1
navneet9431
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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?
 
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  • #2
navneet9431 said:
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##
 
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  • #3
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.
 
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  • #4
fresh_42 said:
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.
 
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  • #5
navneet9431 said:
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.
 
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  • #6
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##
 
  • #7
##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##
PeroK said:
... 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.
 
  • #8
navneet9431 said:
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##.
 
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  • #9
fresh_42 said:
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
$$\lim b=\infty$$
fresh_42 said:
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.
 
  • #10
What does [itex]1\cdot 1\cdot \ldots[/itex] 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.
 
  • #11
For all positive x in ℝ, [itex]\lim\limits_{n\to\infty} x^\frac{1}{n} = 1[/itex].
 
  • #12
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.
 
  • #13
PeroK said:
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!
 
  • #14
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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  • #15
fbs7 said:
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.
 
  • #16
WWGD said:
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):
someone said:
For all positive x in ℝ, [itex]\lim\limits_{n\to\infty} x^\frac{1}{n} = 1[/itex].
 
  • #17
fbs7 said:
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.
 
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  • #18
Matt Benesi said:
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.
 
  • #19
WWGD said:
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 [itex]1^\infty[/itex] was undefined because for all positive x in ℝ [itex]\lim\limits_{n\to\infty} \, x^\frac{1}{n} \, = \, 1[/itex].

So [itex]\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 [/itex] so [itex]\lim\limits_{n\to\infty} \, b^n \, = \, 2 \, = \, 3 \, = \, 4[/itex], etc. unless limits are supposed to be specific, so [itex]\lim\limits_{n\to\infty} \, 2^\frac{1}{n} \, != \,\lim\limits_{n\to\infty} \, 3^\frac{1}{n} \,[/itex].
What about ℂ? [itex] 1 \, = \, ( i^x )^\frac{4}{x} \, = \,e^{i \, \frac{\pi}{2} \cdot x \cdot \frac{4}{x}}[/itex], for all x in ℝ... but what's up with the 0 case?
 
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  • #20
Matt Benesi said:
Ok, I'm not following what you're saying. Can you explain it more?

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

So [itex]\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 [/itex] so [itex]\lim\limits_{n\to\infty} \, b^n \, = \, 2 \, = \, 3 \, = \, 4[/itex], etc. unless limits are supposed to be specific, so [itex]\lim\limits_{n\to\infty} \, 2^\frac{1}{n} \, != \,\lim\limits_{n\to\infty} \, 3^\frac{1}{n} \,[/itex].
What about ℂ? [itex] 1 \, = \, ( i^x )^\frac{4}{x} \, = \,e^{i \, \frac{\pi}{2} \cdot x \cdot \frac{4}{x}}[/itex], for all x in ℝ... but what's 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 ##?
 
  • #21
WWGD said:
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):
WWGD said:
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.
 
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1. What does it mean for a number to be raised to the power of infinity?

Raising a number to the power of infinity means that the number is being multiplied by itself an infinite number of times. This concept is often used in calculus and is known as a limit.

2. Why is raising 1 to the power of infinity considered indeterminate?

When we raise 1 to the power of infinity, the result can vary depending on how we approach the limit. This makes it indeterminate because we cannot determine a single, definite value for the expression.

3. Can we assign a value to 1 to the power of infinity?

No, we cannot assign a specific value to 1 to the power of infinity because it is an indeterminate expression. However, we can use mathematical techniques such as L'Hopital's rule to evaluate the limit and get a more precise value.

4. What are some real-life applications of 1 to the power of infinity?

One real-life application of 1 to the power of infinity is in the study of population growth. When we consider a population growing at a constant rate, we can use the concept of a limit to determine the population size at a specific time in the future.

5. Is 1 to the power of infinity always indeterminate?

No, 1 to the power of infinity is not always indeterminate. It depends on the specific expression and how we approach the limit. In some cases, the limit may converge to a specific value, making the expression determinate.

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