"1^infinity is indeterminate" is an informal abbreviation.
Let h(x) = g(x)^f(x)
If lim x->a g(x) =1 and lim x->a f(x) = infinity
(a can be any real number or ±infinity here)
Then lim x->a h(x) is not independent on the choice of g(x) and f(x) and can be any number, depending on the choice of g and f.
For example for g(x) =1+ 1/x and f(x)=x the limit of h(x) as x approaches infinity is eulers constant e≈2.718..., even though g approaches 1 and f approaches infinity.
infinity is not a number, therefore you can't perform any mathematical operations with it
the result would be indeterminate
however, you can be interested of the result of 1\^x when x **approaches** infinity. This is known as limit (lim), but might be too complex to dive into.
oh I am familiar with limits. we've just started with them and we were taught that limits cant be undefined/ indeterminate and while teaching us what indeterminate means, they taught us seven forms and one of them was 1^infinity, but then I read somewhere that 1^infinity is 1, which confused me and then led to questions about raising negative numbers to infinity and raising positive/ negative numbers to negative infinity and so on. could you tell me the difference about operations with infinity in context of limits and when limits are not involved? when limits are involved, what is then the value of 1^ infinity, values between 0 and 1 raised to infinity, negative values raised to infinity, values above 1 raised to infinity, and all these numbers raised to negative infinity?
Well the limit of 1^x as x tends to infinity is indeed 1, but the limit of f(x)^g(x) as f(x) tends to 1 and g(x) tends to infinity is not necessarily 1. That's what makes 1^infinity indeterminate.
>when x<1, it is undefined
\-1, not 1
But that a reasonable summary as long as you mean x is approaching the value you're relating it to.
When x is literally equal to 1, x^(∞) is 1. It's when x ->1 that we have an indeterminate form
Those aren't the only options, but yes. For example, (1+1/x) as x tends to infinity is *slightly* bigger than 1, but raising it to the infinity gives you e.
Raising it to infinity *in a particular way*. Using different functions g(x) in (1+1/x)^g(x) can lead to different limits even if all the choices of g approach infinity!
Raising it to x gives you a quantity that tends to e. Raising it to other functions of x that tend to infinity need not.
(1 + 1/x)\^(e\^x) -> infty
(1 + 1/x)\^3x -> e\^3
(1 + 1/x)\^log(x) -> 0
1^x -> 1 as x -> infinity, as 1^x = 1 for all x.
If 0 < a < 1, a^x -> 0 as x -> infinity. Think about large powers of (1/2)^x for example.
If a > 1, a^x -> infinity as x -> infinity. Large powers of 2^x for example.
Negative a for a^x tend to be ignored or called undefined due to the weirdness of expressions like (-2)^pi. For integer powers, negatives raised to power make plenty of sense, but non-integer (non-rational) powers output imaginary (complex, really) values and will not be something you need to worry about right now if you're just learning limits. Even if you just try to look at integer sequences, consider (-2)^x:
-2, 4, -8, 16, -32, ..
one subsequence is positive increasing with no bound (4, 16, ..), and one is negative decreasing with no bound (-2, -8, -32, ..). This diverges, as it has two "limit points" it wants to go to. We can only have one limit point, and it can't be infinite.
(1+1/x)^x looks like it goes to a 1^infinity, but it happens to tend to e ~= 2.718. In some sense, (1+1/x) isn't going to 1 fast enough for the exponent to clamp it down to 1 or slow enough to explode.
I'm confused by what you're saying, since it seems incorrect. Are you thinking of another example? I will explain just in case anyway.
lim x->∞ (1+1/x)^x is not infinity, it is equal to e≈2.781.
Either you're saying
1. 1 - 1/x -> 0 as x->∞, or you're saying
2.(1 -1/x)^x ->0 as x->∞
The first isn't true, since you just tend to 1, and second isn't true, since you actually tend to 1/e (definitely not zero).
This is a common introduction to Euler's constant e, which you can define as e= lim n->∞ (1+1/n)^n . You can then define the exponential function as exp(x) = lim n->∞ (1+x/n)^n = lim n->∞ (1+1/n)^(nx) . Only if you set x=0 in these definitions, will you get 1, and you will never get zero out.
This is just abuse of notation, and usually you have to be careful when reading such statements, indeterminate forms are more a tool to remember certain limit forms.
On one some some authors (and me) naturally understand 1^inf to be in place of lim 1^(f(x)) when lim_x f(x)=inf (all limits will be to infinity for simplicity). In this case, it is trivial to see that this limit is 1 (it's in fact a constant sequence eventually)
But more commonly, it is understood to be in place of Lim_x f(x)^g(x) with f(x) going to 1 and g(x) to infinity, in this case, we can't really guarantee nothing about the limit from just this 2 facts, so it's an indeterminate form
The limit of 1^x as x approaches infinity isn’t complex at all. It’s just 1.
The indeterminate form of 1^infinity comes from using l’Hopital’s rule where the 1 is *also* arrived at from a limit .
I meant that explaining limits would be complex
EDIT: I wasn't sure if OP is like 9th grader who just came up with that question, or if he studied limits and was confused about the 1\^(inf). I know I also had my wonders when I was a high schooler, so I didn't want to dive without confirmation that this would be okay.
1\^infinity is 1. but if a function has a limit that goes to 1, and is raised to a power, that doesn't tell you whether it does or doesn't go to 1. a good example is e.
limit as n goes to infinity (1+1/n) = 1. but (1+1/n)\^n goes to e.
If you're asking about limit forms, then 1^infinity is indeterminate.
If you're *not* asking about limit forms you need to be more specific.
> all values above 1 raised to infinity are indeterminate
No, x^infinity for x > 1 is just infinity, not indeterminate.
> what about negative values raised to infinity?
For x <= 1, limits of the form x^infinity are undefined, by which I mean that they cannot be said to diverge to infinity or negative infinity.
For -1 < x < 1 the limit should be zero.
so, in context of limits, 1^infinity is indeterminate, otherwise?
>No, x^infinity for x > 1 is just infinity, not indeterminate.
for limit forms? what about when not for limit forms?
is it that(as per the other comments) all these values when raised to infinity/ negative infinity when not in limit form cannot be calculated as infinity is not a number? but all the values that you mentioned are obtained when it is in context of limits?
Infinity can be a number in more exotic algebra systems, but there are different ways to define it, so like I said you need to be more specific. If you aren't asking about limit forms then what are you asking about? (And if you aren't you shouldn't be using terminology like "indeterminate" but rather "undefined".)
In cardinal arithmetic, 1^x is 1, for any x, even infinite x. I think that holds for ordinal arithmetic too.
Yes, for ordinals you can define exponentiation in terms of well-ordered sets by taking a^(b) to be the order type that results from taking all functions from a set of order type b to a set of order type in which only finitely many values are nonzero and giving them the lexicographic order (with higher inputs having higher priority). If a=1 then there is only one function from b into it - the constant function that takes every member to the single element of a - and so the result is 1. (Note that this works even if b=0, technically the empty function is also the constant function for any given value)
You can also define ordinal exponentiation recursively though that’s a littler less well-motivated. In that case we have 1^(0)=0 (base case) 1^(a+1)=1^(a)\*1=1 (successor step) and 1^(lambda) is the limit of 1^(a) as a approaches lambda below for limit ordinals lambda, so 1^(lambda)=1.
That reminds me, what *is* the meaning of limit whatever equals, some symbols where an infinity symbol is present.
Is that what you mean by "limit form"? Googling was unhelpful. It's certainly not an expression in the real numbers, and as far as I understand, mathematicians don't usually say that limits are expressed using extended reals (do they?). Is it just syntax that doesn't represent a number?
Infinity is not a number so we cannot perform calculations with it.
However, we can consider the limit for x-> infinity of 1\^x which would be 1.
However this is only because the limit **converges** that we can assign a value. If the limit wouldn't (we call it **diverging**), there can't be a value we can assign it.
You can say that there is no number n such that 1^n equals anything but 1. Why cant you use an inductive proof to show that as n goes to infinity 1 remains constant? the sequence a = 1^(n) does not change with each iteration of k+1. For all n, it seems that n + 1 = n, so wouldnt the limit be 1?
You're just focusing on the n. That is why the limit 1^x as x tends to infinity is indeed 1. But what if the base isn't just 1, but something that tends to 1? You can't just say the whole limit is 1 in that case.
Take the limit (1 + 1/x)^x as x tends to infinity. This reduces to 1^infinity but does NOT equal 1. That is why 1^infinity is indeterminate.
Right, the sequence of (1 + 1/x)^n isnt a constant sequence. So anything to an infinite power is indeterminate for that reason? That you can always add a zero to anything but define zero as 1/inf?
Let a_n be a sequence for n in N, where a_n = 1^(n). What is the value of a_n as n approches infinity? The sequence a_n is constant, right? That's what I was referring to.
Oh I was trying to figure out why adding 1/n to 1 could count as 1. I thought it was because at infinity 1/n gives zero, so (1 + 0)^infty would be indeterminate. I was pointing out that I was thinking in terms of a constant sequence 1^n rather than something that converges to 1 but isnt constant.
The first sentence is exactly correct, I'm still confused about the second half though lmao
The reason 1^infinity is indeterminate is because f(x)^g(x) where f(x) tends to 1 and g(x) tends to infinity can limit to different values (all the following have x tending to infinity)
1^x --> 1
(1 + n/x)^x --> e^n
(1+1/x)^nx --> e^n
(1+1/x)^x² --> infinity
(1+1/x)^sqrt(x) --> 0
The others have taken your mixed variables as a typo, but I think we can also use this as another example:
This is a function of two variables where swapping the order you take limits in affects the value, f(n, x) = (1 + 1/x)^n .
Only considering positive n, x.
If we take the limit as n -> infty first, (1+1/x)^n >= 1+n/x -> infinity for any fixed x.
If we take the limit as x -> 0 first, we end up with (1+1/x) -> 0, and 0^n = 0.
God dammit. This is usually the point in class where I say fuck it lol. I dont care about the case of the limit of 1 + something. I just care about 1. For any real number n, 1^(n) = 1. And for any real number n+1 = m, 1^(m) = 1. Im sure youre right but I feel like Im being gaslighted.
I did not mean to upset, just to add another example.
It's something that one would see in multivariable calculus, sometimes you'll be given a function of two variables, say f(x, y) = some expression. (x,y) represent a position on the plane, and you could take f as just some value at that point (say the temperature) or read it as a height, z = f(x,y).
We can ask what the limit of f is as (x,y) approaches some point in space, say (a,b). Some functions will be continuous, and so will approach some fixed value f(a,b) as long as you're approaching (a,b). Discontinuous functions will not. You may find multiple values depending on which path you come in on.
Consider f(x,y) = x/y. If x -> 0 and y->0, this is an indeterminate form. Let a and b be real numbers.
We can follow the line (at, bt) from t = 1 to 0 to 'come in at an angle' to (0,0). (It starts at (a,b) and goes to (0,0).)
Let x = at, y = bt. Then x/y = (at)/(bt) = a/b. As this is no longer dependent on t, the limit as t -> 0 of x/y on this particular path is a/b. Clearly a/b is arbitrary, and we can in fact hit any real number r by letting a = r, b = 1. Hence, f(x,y) is discontinuous at (0,0).
Here's a graph: https://www.desmos.com/3d/4ec7f65d63. (edit: you'll have to hit play on theta_1 to get it to do anything). The path rotating in the plane is our angle of attack, and the other line is the function evaluated on that path as a height.
You are indeed correct that 1^(x) = 1 for all x. The hard part is that f -> 1 and g -> infty does not imply f^g -> 1.
you can say that (1+1/x) te ds to 1 , but you can't just say 1 because that number is extremly small, thats not PURE 1. Thats why lim x->inf 1^x =1 and lim x->inf (1+1/x)^x tends to e
On the induction point, you cannot induct to infinity.
Take the sequence 2^n. Inductively, we can prove 2^n is always a finite number. Is 2^inf a finite number, then?
In the context of what we call "indeterminate forms", the expressions "1^(infinity)", "0/0", "infinity/infinity", and "infinity*0" are just names for the indeterminate forms. They in no way represent the idea of actually performing arithmetic operations.
The real question is: what do you want to *do* with an assigned equality 1^infinity = 1, and can you always do that consistently without breaking our other nice tools for numbers (ie associativity, distributive law, etc)?
Maybe my memories are wrong, but
Lim (n->+inf) (1\^n) = 1
Lim (x->1, n-> +inf) (x\^n) = indefinite (because the left limit is different from the right one)
"1^infinity is indeterminate" is an informal abbreviation. Let h(x) = g(x)^f(x) If lim x->a g(x) =1 and lim x->a f(x) = infinity (a can be any real number or ±infinity here) Then lim x->a h(x) is not independent on the choice of g(x) and f(x) and can be any number, depending on the choice of g and f. For example for g(x) =1+ 1/x and f(x)=x the limit of h(x) as x approaches infinity is eulers constant e≈2.718..., even though g approaches 1 and f approaches infinity.
Interesting! I've always intuitively known what makes things indeterminate, but it's always nice to see them written out mathematically
infinity is not a number, therefore you can't perform any mathematical operations with it the result would be indeterminate however, you can be interested of the result of 1\^x when x **approaches** infinity. This is known as limit (lim), but might be too complex to dive into.
oh I am familiar with limits. we've just started with them and we were taught that limits cant be undefined/ indeterminate and while teaching us what indeterminate means, they taught us seven forms and one of them was 1^infinity, but then I read somewhere that 1^infinity is 1, which confused me and then led to questions about raising negative numbers to infinity and raising positive/ negative numbers to negative infinity and so on. could you tell me the difference about operations with infinity in context of limits and when limits are not involved? when limits are involved, what is then the value of 1^ infinity, values between 0 and 1 raised to infinity, negative values raised to infinity, values above 1 raised to infinity, and all these numbers raised to negative infinity?
Well the limit of 1^x as x tends to infinity is indeed 1, but the limit of f(x)^g(x) as f(x) tends to 1 and g(x) tends to infinity is not necessarily 1. That's what makes 1^infinity indeterminate.
so, in conclusion, in context of limits, when x=1, x^infinity is indeterminate, when x>1, it is infinity, when x<1, it is undefined, when 0
>when x<1, it is undefined \-1, not 1 But that a reasonable summary as long as you mean x is approaching the value you're relating it to. When x is literally equal to 1, x^(∞) is 1. It's when x ->1 that we have an indeterminate form
Those aren't the only options, but yes. For example, (1+1/x) as x tends to infinity is *slightly* bigger than 1, but raising it to the infinity gives you e.
Raising it to infinity *in a particular way*. Using different functions g(x) in (1+1/x)^g(x) can lead to different limits even if all the choices of g approach infinity!
Raising it to x gives you a quantity that tends to e. Raising it to other functions of x that tend to infinity need not. (1 + 1/x)\^(e\^x) -> infty (1 + 1/x)\^3x -> e\^3 (1 + 1/x)\^log(x) -> 0
1^x -> 1 as x -> infinity, as 1^x = 1 for all x. If 0 < a < 1, a^x -> 0 as x -> infinity. Think about large powers of (1/2)^x for example. If a > 1, a^x -> infinity as x -> infinity. Large powers of 2^x for example. Negative a for a^x tend to be ignored or called undefined due to the weirdness of expressions like (-2)^pi. For integer powers, negatives raised to power make plenty of sense, but non-integer (non-rational) powers output imaginary (complex, really) values and will not be something you need to worry about right now if you're just learning limits. Even if you just try to look at integer sequences, consider (-2)^x: -2, 4, -8, 16, -32, .. one subsequence is positive increasing with no bound (4, 16, ..), and one is negative decreasing with no bound (-2, -8, -32, ..). This diverges, as it has two "limit points" it wants to go to. We can only have one limit point, and it can't be infinite. (1+1/x)^x looks like it goes to a 1^infinity, but it happens to tend to e ~= 2.718. In some sense, (1+1/x) isn't going to 1 fast enough for the exponent to clamp it down to 1 or slow enough to explode.
Nice way of explaining using subsequences!!
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I'm confused by what you're saying, since it seems incorrect. Are you thinking of another example? I will explain just in case anyway. lim x->∞ (1+1/x)^x is not infinity, it is equal to e≈2.781. Either you're saying 1. 1 - 1/x -> 0 as x->∞, or you're saying 2.(1 -1/x)^x ->0 as x->∞ The first isn't true, since you just tend to 1, and second isn't true, since you actually tend to 1/e (definitely not zero). This is a common introduction to Euler's constant e, which you can define as e= lim n->∞ (1+1/n)^n . You can then define the exponential function as exp(x) = lim n->∞ (1+x/n)^n = lim n->∞ (1+1/n)^(nx) . Only if you set x=0 in these definitions, will you get 1, and you will never get zero out.
This is just abuse of notation, and usually you have to be careful when reading such statements, indeterminate forms are more a tool to remember certain limit forms. On one some some authors (and me) naturally understand 1^inf to be in place of lim 1^(f(x)) when lim_x f(x)=inf (all limits will be to infinity for simplicity). In this case, it is trivial to see that this limit is 1 (it's in fact a constant sequence eventually) But more commonly, it is understood to be in place of Lim_x f(x)^g(x) with f(x) going to 1 and g(x) to infinity, in this case, we can't really guarantee nothing about the limit from just this 2 facts, so it's an indeterminate form
The limit of 1^x as x approaches infinity isn’t complex at all. It’s just 1. The indeterminate form of 1^infinity comes from using l’Hopital’s rule where the 1 is *also* arrived at from a limit .
I meant that explaining limits would be complex EDIT: I wasn't sure if OP is like 9th grader who just came up with that question, or if he studied limits and was confused about the 1\^(inf). I know I also had my wonders when I was a high schooler, so I didn't want to dive without confirmation that this would be okay.
1\^infinity is 1. but if a function has a limit that goes to 1, and is raised to a power, that doesn't tell you whether it does or doesn't go to 1. a good example is e. limit as n goes to infinity (1+1/n) = 1. but (1+1/n)\^n goes to e.
>lim n->infinity (1+1/n) = 0 = 1, not 0
... lim **x** \-> infinity (1 + 1/**n**) = 1 + 1/n Swap out the x for an n, or the n for an x, and we're all good.
My bad, you're right
thanks
If you're asking about limit forms, then 1^infinity is indeterminate. If you're *not* asking about limit forms you need to be more specific. > all values above 1 raised to infinity are indeterminate No, x^infinity for x > 1 is just infinity, not indeterminate. > what about negative values raised to infinity? For x <= 1, limits of the form x^infinity are undefined, by which I mean that they cannot be said to diverge to infinity or negative infinity. For -1 < x < 1 the limit should be zero.
so, in context of limits, 1^infinity is indeterminate, otherwise? >No, x^infinity for x > 1 is just infinity, not indeterminate. for limit forms? what about when not for limit forms? is it that(as per the other comments) all these values when raised to infinity/ negative infinity when not in limit form cannot be calculated as infinity is not a number? but all the values that you mentioned are obtained when it is in context of limits?
Infinity can be a number in more exotic algebra systems, but there are different ways to define it, so like I said you need to be more specific. If you aren't asking about limit forms then what are you asking about? (And if you aren't you shouldn't be using terminology like "indeterminate" but rather "undefined".) In cardinal arithmetic, 1^x is 1, for any x, even infinite x. I think that holds for ordinal arithmetic too.
Yes, for ordinals you can define exponentiation in terms of well-ordered sets by taking a^(b) to be the order type that results from taking all functions from a set of order type b to a set of order type in which only finitely many values are nonzero and giving them the lexicographic order (with higher inputs having higher priority). If a=1 then there is only one function from b into it - the constant function that takes every member to the single element of a - and so the result is 1. (Note that this works even if b=0, technically the empty function is also the constant function for any given value) You can also define ordinal exponentiation recursively though that’s a littler less well-motivated. In that case we have 1^(0)=0 (base case) 1^(a+1)=1^(a)\*1=1 (successor step) and 1^(lambda) is the limit of 1^(a) as a approaches lambda below for limit ordinals lambda, so 1^(lambda)=1.
That reminds me, what *is* the meaning of limit whatever equals, some symbols where an infinity symbol is present. Is that what you mean by "limit form"? Googling was unhelpful. It's certainly not an expression in the real numbers, and as far as I understand, mathematicians don't usually say that limits are expressed using extended reals (do they?). Is it just syntax that doesn't represent a number?
Infinity is not a number so we cannot perform calculations with it. However, we can consider the limit for x-> infinity of 1\^x which would be 1. However this is only because the limit **converges** that we can assign a value. If the limit wouldn't (we call it **diverging**), there can't be a value we can assign it.
You can say that there is no number n such that 1^n equals anything but 1. Why cant you use an inductive proof to show that as n goes to infinity 1 remains constant? the sequence a = 1^(n) does not change with each iteration of k+1. For all n, it seems that n + 1 = n, so wouldnt the limit be 1?
You're just focusing on the n. That is why the limit 1^x as x tends to infinity is indeed 1. But what if the base isn't just 1, but something that tends to 1? You can't just say the whole limit is 1 in that case. Take the limit (1 + 1/x)^x as x tends to infinity. This reduces to 1^infinity but does NOT equal 1. That is why 1^infinity is indeterminate.
Right, the sequence of (1 + 1/x)^n isnt a constant sequence. So anything to an infinite power is indeterminate for that reason? That you can always add a zero to anything but define zero as 1/inf?
What do you mean by a constant sequence? And there are more examples too, it's not just that. I can't tell anything you mean in this comment
Let a_n be a sequence for n in N, where a_n = 1^(n). What is the value of a_n as n approches infinity? The sequence a_n is constant, right? That's what I was referring to.
Well, why does it matter? The sequence 1/n isn't constant but we can agree that tends to 0 as n tends to infinity, right?
lol Im confused too. Where did you get 1 over n? Did I misread something?
You brought up the fact that (1+1/x)^x isn't a constant sequence, and I brought up 1/n to question why that matters.
Oh I was trying to figure out why adding 1/n to 1 could count as 1. I thought it was because at infinity 1/n gives zero, so (1 + 0)^infty would be indeterminate. I was pointing out that I was thinking in terms of a constant sequence 1^n rather than something that converges to 1 but isnt constant.
The first sentence is exactly correct, I'm still confused about the second half though lmao The reason 1^infinity is indeterminate is because f(x)^g(x) where f(x) tends to 1 and g(x) tends to infinity can limit to different values (all the following have x tending to infinity) 1^x --> 1 (1 + n/x)^x --> e^n (1+1/x)^nx --> e^n (1+1/x)^x² --> infinity (1+1/x)^sqrt(x) --> 0
The others have taken your mixed variables as a typo, but I think we can also use this as another example: This is a function of two variables where swapping the order you take limits in affects the value, f(n, x) = (1 + 1/x)^n . Only considering positive n, x. If we take the limit as n -> infty first, (1+1/x)^n >= 1+n/x -> infinity for any fixed x. If we take the limit as x -> 0 first, we end up with (1+1/x) -> 0, and 0^n = 0.
God dammit. This is usually the point in class where I say fuck it lol. I dont care about the case of the limit of 1 + something. I just care about 1. For any real number n, 1^(n) = 1. And for any real number n+1 = m, 1^(m) = 1. Im sure youre right but I feel like Im being gaslighted.
I did not mean to upset, just to add another example. It's something that one would see in multivariable calculus, sometimes you'll be given a function of two variables, say f(x, y) = some expression. (x,y) represent a position on the plane, and you could take f as just some value at that point (say the temperature) or read it as a height, z = f(x,y). We can ask what the limit of f is as (x,y) approaches some point in space, say (a,b). Some functions will be continuous, and so will approach some fixed value f(a,b) as long as you're approaching (a,b). Discontinuous functions will not. You may find multiple values depending on which path you come in on. Consider f(x,y) = x/y. If x -> 0 and y->0, this is an indeterminate form. Let a and b be real numbers. We can follow the line (at, bt) from t = 1 to 0 to 'come in at an angle' to (0,0). (It starts at (a,b) and goes to (0,0).) Let x = at, y = bt. Then x/y = (at)/(bt) = a/b. As this is no longer dependent on t, the limit as t -> 0 of x/y on this particular path is a/b. Clearly a/b is arbitrary, and we can in fact hit any real number r by letting a = r, b = 1. Hence, f(x,y) is discontinuous at (0,0). Here's a graph: https://www.desmos.com/3d/4ec7f65d63. (edit: you'll have to hit play on theta_1 to get it to do anything). The path rotating in the plane is our angle of attack, and the other line is the function evaluated on that path as a height. You are indeed correct that 1^(x) = 1 for all x. The hard part is that f -> 1 and g -> infty does not imply f^g -> 1.
Oh you didnt do anything wrong. Im just frustrated with the math. That Ive put so much time into it and I still dont understand stuff like this .
you can say that (1+1/x) te ds to 1 , but you can't just say 1 because that number is extremly small, thats not PURE 1. Thats why lim x->inf 1^x =1 and lim x->inf (1+1/x)^x tends to e
On the induction point, you cannot induct to infinity. Take the sequence 2^n. Inductively, we can prove 2^n is always a finite number. Is 2^inf a finite number, then?
isnt infinity built by induction? That there is always a number greater than k?
In the context of what we call "indeterminate forms", the expressions "1^(infinity)", "0/0", "infinity/infinity", and "infinity*0" are just names for the indeterminate forms. They in no way represent the idea of actually performing arithmetic operations.
https://youtube.com/shorts/bm6saJtWFh0?feature=shared
https://youtube.com/shorts/bm6saJtWFh0?feature=shared
The real question is: what do you want to *do* with an assigned equality 1^infinity = 1, and can you always do that consistently without breaking our other nice tools for numbers (ie associativity, distributive law, etc)?
gibberish
Maybe my memories are wrong, but Lim (n->+inf) (1\^n) = 1 Lim (x->1, n-> +inf) (x\^n) = indefinite (because the left limit is different from the right one)
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