# 1 raised to Infinity becomes finite???

Evaluate $1^{\infty}$.

Most of the students if asked this question will answer 1, since $1^{\infty}$ is seems to be equal to

($1\times 1\times 1\times \ldots\times 1)$ with one being multiplied by itself infinite times.

However, more advance mathematics introduces us principle of limits and able to evaluate like cases.

Now, consider this limit

$\lim_{x \to 0} (1+x)^{1/x}$ ,

and we all know the limit gives us constant $e$.

Consequently, as $x \to 0$, $1+x \to 1$ and $1/x \to \infty$, so the limit will give us the limit of $1^{\infty}$.

Questions:

a) Is $1^{\infty}$ a finite number?(or somewhat equal to $e$ as what the arguments stated above)

b) Are there errors in my understandings particularly in my arguments being used?

c)Resolve the argument or give proof/disproof that $1^{\infty}$ is not equal to one.

Thanks for responding. I am really confused about this.

Note by Mharfe Micaroz
6 years, 4 months ago

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## Comments

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To know whether 1 ^ ∞ = 1 or not, we need to understand the meaning of We all know that _ ∞ _ is not defined, but why is it so ?

Rather it should be defined (according to me). Now I tell you something, try it, do it : Ask yourselves a few questions -

• How many points are there on a line segment of length 2 cm ?
• How many points are there on a line segment of length 10 cm ?

Of course your answers will be the same, but do you think it should be the same i.e. there are same no. of points on line segment of 2 cm and that of 10 cm but then what's the point of difference between them ??

- 6 years, 4 months ago

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1^∞ is an indeterminate term because for limit we take left hand(lhl) as well right hand limit(rhl). For this case lhl is 0 as 1-h where h tends 0 is term smaller than 1 and on repeated multiplication will go on to 0 whereas rhl is ∞ because 1+h where h tends 0 is greater than 1 and on repeated multiplication goes to ∞....

- 6 years, 4 months ago

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please explain a bit clear even i have the same doubt

pls!!

- 6 years, 4 months ago

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First of all, $\infty$ is not a real or a complex number, so you should not appeal to properties of those number systems to evaluate expressions with infinity. You can sometimes study expressions with infinity by interpreting it as a limit, but that is a human reinterpretation of the question.

Even if this question were interpreted as a limit, the most direct translation would be to: $\lim_{x \to \infty} 1^x$ And this limit does equal 1.

A less direct interpretation would be to interpret it as: $\lim_{a \to 1 \text{and} x \to \infty} a^x$ This expression is undefined unless there's a relationship between a and x.

There are some number systems (such as the surreal numbers) which do include arithmetic properties of infinity, but you will have to be clear about what number system you intend, and understand the properties of that number system when evaluating the expression. In the surreal number system, I believe $1^{\omega}$ evaluates to 1.

- 6 years, 4 months ago

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First of all, $1^{\infty}$, in this case, is not a number. It is what mathematicians call an indeterminate form (because it can't be uniquely determined what it is).

It could be represented by something of the form $\lim_{x \to a} {f(x)^{g(x)}}$, where $f(x)$ and $g(x)$, are functions such that when $x \to {a}$ (for $a \in \mathbb{R}$ or $a = \pm \infty$), $f(x) \to 1$ and $g(x) \to \infty$.

One of the standard ways in which its value can be resolved (and in general, whether it is convergent or divergent) is to convert it into an indeterminate form of $\frac{0}{0}$ or $\frac{\infty}{\infty}$ by doing this neat trick:

$\lim_{x \to a} {f(x)^{g(x)}}$ = $\exp(\lim_{x \to a} {\frac{\ln{f(x)}}{1/g(x)}})$

or

$\lim_{x \to a} {f(x)^{g(x)}}$ = $\exp(\lim_{x \to a} {\frac{g(x)}{1/\ln{f(x)}}})$

These limits can then be standardly evaluated using l'Hôpital's rule. For example, in the case of $\lim_{x \to 0} {(1+x)^{1/x}}$:

$\lim_{x \to 0} {(1+x)^{1/x}}$ = $\exp(\lim_{x \to 0} {\frac{\ln{(1+x)}}{x}})$ = $\exp(\lim_{x \to 0} {\frac{1/(1+x)}{1}})$ (using l'Hôpital's rule) = $\exp(1)$ = $e$.

Of course, this all changes if by 1 you mean the constant function $f(x) = 1$. I think that in that case this indeterminate form evaluates to 1 every time.

- 6 years, 4 months ago

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You mean this trick is for intederminate form of $1^\infty$ or $\infty^0$.

- 6 years, 4 months ago

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Yes. It also works for $0^0$.

- 6 years, 4 months ago

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1 raised to any power is 1, always.

- 6 years, 4 months ago

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Hello :)

- 6 years, 4 months ago

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I think everything has $1^{\infty}$ as their identity.

Look at my example below.

If we have number $2$. Then we write like this $2 \times 1$. After that, we also write like this $2 \times 1 \times 1$, like this too $1 \times 2 \times 1 \times 1$ and so on or we simplify like this $2 \times 1^{\infty}$ and like this $1^{\infty} \times 2 \times 1^{\infty}$ and so on again.

isn't it?

- 6 years, 4 months ago

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u on skype??

- 6 years, 4 months ago

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Yes, I still trying to be present everywhere.

What's the matter?

- 6 years, 4 months ago

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We already discussed this a week and a half ago. It shouldn't be too far down: Page 3 or so. read that.

- 6 years, 4 months ago

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am i forbidden here to ask this genius people?..

- 6 years, 4 months ago

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$\lim_{n \rightarrow \infty} 1^n = 1$ but if you are looking for $\lim_{n \rightarrow \infty} f(n)^{g(n)}$ where $\lim_{n \rightarrow \infty} f(n) = 1$ and $\lim_{n \rightarrow \infty} g(n) = \infty$, it depends on the specific $f$ and $g$.

- 6 years, 4 months ago

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On limit studies, $1^\infty$ is not determined.

I think that happens because this limit depends on who approaches "1" faster: the basis or the exponent. It may result 0 if the exponent is much faster and the basis goes to 1 from above. It may result 0 if the exponent is faster and the basis approaches from below...

So...

a) It is not a number. In fact, $\infty$ is not a real number (maybe an extended one), so the only way to define $1^\infty$ is using limit process.

b) The error is not in the arguments, but in it's implications for calculus. We just can't define this limit properly.

c) You can find some exemples of limits giving indetermination $1^\infty$ and in fact may result 0, $\infty$, 1...

- 6 years, 4 months ago

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Victor, I want to ask you something.

what is the definition of infinity ($\infty$)?

is $\infty$ the max number or the greatest one? or is $\infty$ something which just extend and neverending? or maybe you have another definition for $\infty$?

- 6 years, 4 months ago

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Many philosophers have made this same question. My nonlinear programming professor said that $\infty$ is a number, such is bigger than any number you can think. Sorry for my grammar.

- 6 years, 4 months ago

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Actually infinity cannot be regarded a number nor any quantity, according to my point of view, rather it is a variable whose value depends upon the conditions where it is used.

- 6 years, 4 months ago

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variable?

hmmm. do you mean that infinity is indeterminate?

indeterminate can be anything, but infinity is just infinity. I think they are different.

hmmm.

ah, yes I know. this webpage

- 6 years, 4 months ago

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No I don't mean that infinity is indeterminate but something whose value depends upon the situation where it is used (as I sated earlier !). In short I mean it's value can be determined but is variable.

- 6 years, 4 months ago

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if $\infty$ is a number, then maybe there are $-\infty$ and $+\infty$ too.

now draw a circle (for Cartesian proofing) or a sphere (for 3D proofing) and assume the center of the circle or sphere is zero ($0$), also assume the perimeter of the circle or the surface of the sphere is infinity ($\infty$). that shape like the Expanding Universe Theory.

it's okay Francilio, grammar isn't important if the other know what you mean. :)

- 6 years, 4 months ago

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One way to prove the statement wrong is to assume for some time that 1^infinity is equal to some finite number say "k". Now, putting this in expression form, 1^infinity=k , now doing log(to the base 1) on both sides, we will get, infinity = log k, as log k is equal to k which is a finite number, it will lead to contradiction as infinity> any finite number and cannot be equal. So it proves the statement that 1^infinite is either not definite or is either 1. Note:-All the logs have been taken with the base 1.

- 6 years, 4 months ago

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no....this is wrong........log with base 1 is not defined

- 6 years, 4 months ago

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Log with base one ins not well defined.

If we could do log(to the base 1) on both sides we could do this: $1^2 = 1^3 \Rightarrow 2=3$

- 6 years, 4 months ago

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but he can make $1=e^0$ and use the neperian logarithm. It results $0=k$. The problem is, math barely can handle with $\infty$ or $0$ (and their correlates) without making some paradoxes, as you can see easily in the barber paradox or the halting problem

- 6 years, 4 months ago

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Thanks, Victor.

Now I have new knowledge. ^_^

- 6 years, 4 months ago

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Sorry, typo: " It may result $\infty$ if the exponent is much faster and the basis goes to 1 from above."

- 6 years, 4 months ago

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you know, you can edit your posts.

- 6 years, 4 months ago

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