How are [math]0\cdot\infty[/math] and [math]1^\infty[/math] "indeterminate forms", given that [math]0+0+0...=0[/math] and [math]1*1*1...=1[/math]?

[math]0[/math] is the identity element of addition, and no matter how many you add together, it will still be [math]0[/math]. The same goes for [math]1[/math] and multiplication. What gives?

Because there are sequences that do not converge to 0 or 1. Algebraic rules do not apply here.

Sebastian Smith

Because arithmetic does not apply to infinity

John Ward

you are right, actually. if you define arithmetic with infinity, it works like that

however, 0*inf and 1^inf usually don't mean arithmetic. they usually mean babby calculus student tried plug and chug and got shit, and now he has to fix it

see for example, talking about convergence when it has nothing to do here

Joseph Butler

[math]0 \times your^{mom}[/math]

[math]1^{{your}^{mom}}[/math]

Why are these undefined?

Henry Adams

So, exactly how and at what point can adding zeros become something else than zero, or multiplying ones become something else then one? Would this not be in direct contradiction of their definitions as identity elements of addition and multiplication respectively? That's like saying that [math]\displaystyle \int_{0}^{\infty}0\,\mathrm{d}x[/math] might be something else than [math]0[/math] because "hurr it goes to infinity and who knows".

Julian Hill

yeah, it does. again, "indeterminate forms" are people misusing "plug and chug" laws

Jace Lee

Take the x -> 0 limit of x/x. That has the form [math]0 \cdot \infty[/math], since it's the same as x times 1/x. This limit is equal to 1. Do the same for x^2/x, and x/x^2. They also have the form [math]0 \cdot \infty[/math], but the limits here are 0 and infinity, respectively. You can do the same for of thing with exponentiation.

Nathan Campbell

finals are coming up

why not stop worrying about why they're indeterminate and start trying to learn how to solve indeterminite forms, nii-san?

Nathaniel Cooper

>Because arithmetic does not apply to infinity

But the operations of adding zeros or multiplying ones are entirely predictable, continuing them ad infinitum does not change their nature in the slightest. It's like saying that you might eventually squeeze some water out of a dry rock if you keep squeezing it indefinitely. Similarly, we know that a definite integral of a constant function is zero even if one of the bounds is infinite, because we know a constant function isn't going to increase nor decrease _anywhere_.

Adam Powell

No, arithmetic works perfectly fine in this cases. $0.\infty = 0$ and $1^{\infty} = 1$. However, in $$\lim_{x\to0} x \times \frac{1}{x}$$, you should get $0.\infty$, but the number in the left isn't quite 0, it's a reaaally small number that's very close to 0, so, the multiplication could result in any real number, because, in the right, you have a reaaally large number. In this case ($x\times \frac{1}{x}$), the result is $1$. Next case: $$\lim_{x\to1} x^{\infty}$$. In this case, x isn't quite $1$, but a number very close to it. So, you can't say that the result is 1, because $x$ is elevated to a really large number. If $x$ is very close but smaller than 1 ($$\lim_{x\to1^{-}}$$), the result is 0 (any number between 0 and 1 elevated to a really large number tends to 0). In the other case, if $x$ is greater than 1, the result is $\infty$ (any number grater than 1 elevated to a large number tends to $\infty$). That's it, sorry for the large text.

William Johnson

>the number on the left isn't quite 0 what, it's 0.00...01 or some ridiculous shit like that?

misusing plug and chug laws gives you wrong shit. 0 means 0.

Adam Bailey

Correction: No, arithmetic works perfectly fine in this cases. [math]0.\infty = 0[\math] and [math]1^{\infty} = 1[\math]. However, in [math]\lim_{x\to0} x \times \frac{1}{x}[\math], you should get [math]0.\infty[\math], but the number in the left isn't quite 0, it's a reaaally small number that's very close to 0, so, the multiplication could result in any real number, because, in the right, you have a reaaally large number. In this case ([math]x\times \frac{1}{x}[\math]), the result is 1. Next case: [math]\lim_{x\to1} x^{\infty}[\math]. In this case, x isn't quite 1, but a number very close to it. So, you can't say that the result is 1, because $x$ is elevated to a really large number. If x is very close but smaller than 1 ([math]\lim_{x\to1^{-}}[\math]), the result is 0 (any number between 0 and 1 elevated to a really large number tends to 0). In the other case, if x is greater than 1, the result is [math]\infty[\math] (any number grater than 1 elevated to a large number tends to [math]\infty[\math]). That's it, sorry for the large text.

Ayden Jenkins

I really don't know how to write in Tex here. Sorry.

Hunter Rogers

>That has the form 0⋅∞, since it's the same as x times 1/x

That is based on the preassumption that any real number divided by [math]\infty[/math] is [math]0[/math], which in turn relies on the assumption that [math]0\cdot\infty[/math] can equal any real number. It's a circular argument really.

Brody Taylor

Yeah, it's some ridiculous shit like 0.00...01. Because that 0 is not 0, it's a number tending to 0.

Easton Morgan

No, arithmetic works perfectly fine in this cases. [math]0\.\infty = 0[/math] and [math]1^{\infty} = 1[/math]. However, in [math]\lim_{x\to0} x \times \frac{1}{x}[/math], you should get [math]0.\infty[/math], but the number in the left isn't quite [math]0[/math], it's a reaaally small number that's very close to [math]0[/math], so, the multiplication could result in any real number, because, in the right, you have a reaaally large number. In this case ([math]x\times \frac{1}{x}[/math]), the result is [math]1[/math]. Next case: [math]\lim_{x\to1} x^{\infty}[/math]. In this case, [math]x[/math] isn't quite [math]1[/math], but a number very close to it. So, you can't say that the result is [math]1[/math], because [math]x[/math] is elevated to a really large number. If [math]x[/math] is very close but smaller than [math]1[/math] ([math]\lim_{x\to1^{-}}[/math]), the result is [math]0[/math] (any number between [math]0[/math] and [math]1[/math] elevated to a really large number tends to [math]0[/math]). In the other case, if [math]x[/math] is greater than [math]1[/math], the result is [math]\infty[/math] (any number grater than [math]1[/math] elevated to a large number tends to [math]\infty[/math]). That's it, sorry for the large text.

Here you go bud, don't worry.

Jose Jenkins

>\. You mean \cdot?

Liam Baker

yus woops

Joseph Jackson

No, that's not based on the assumption that [math]0.\infty[/math] can equal any real number. It's the opposite. The fact that [math]0.\infty[/math] can equal any real number comes from tha fact that any real number divided by another that is tending to [math]\infty[/math] is equal to 0. And look, [math]\infty[/math] is not a number like 2 or 3, just to be clear. A tendency to [math]\infty[/math] can occur in multiple manners, fast, slow, etc.

Lucas Mitchell

Thanks, man! I was putting the wrong bar in /math...

The fact that [math]\displaystyle \frac{0}{0}[/math] is indeterminate is consistent though, as opposed to the expressions brought up by OP. This can be shown as follows:

1) [math]0[/math] is the identity element of addition. 2) Therefore, [math]0+0+0...=0[/math] for any number [math]x[/math] of summands, EVEN IF the summands continue indefinitely, i.e if their number is unbounded, i.e. if [math]x=\infty[/math] - this is an inevitable consequence of 1). 3) Therefore, [math]x \cdot 0 = 0[/math] for any [math]x \in \mathbb{R}\cup\{\infty\}[/math]. 4) Therefore, [math]\displaystyle \frac{0}{0}[/math] can be satisfied by any number from [math]\mathbb{R}\cup\{\infty\}[/math], and is therefore indeterminate. 5) Therefore, for any number [math]a \neq 0[/math], the expression [math]\displaystyle \frac{a}{0}[/math] is entirely meaningless (i.e. undefined), because [math]x \cdot 0 \neq a[/math] for any [math]x \in \mathbb{R}\cup\{\infty\}[/math]. For the same reasons, the expression [math]\displaystyle \frac{a}{\infty}[/math] would represent an infinitesimally small quantity, but never [math]0[/math] itself (as the reverse operation [math]0 \cdot \infty[/math] could never equal an [math]a \neq 0[/math]).

Referring back to analogy, defining [math]0[/math] as the identity element of addition is like defining a certain rock to be entirely void of any water, while expecting [math]0 \cdot \infty[/math] to yield anything else than [math]0[/math] is like expecting of said entirely dry rock to somehow shed water anyway if sqeezed ad infinitum.

Ian Watson

>Do you know what indeterminate means? >If the limit of some combination of functions depends on the functions selected and not just the value of the functions in the limit then it is an indeterminate form. >For example lim x->0 f(x)/g(x) where lim x->0 f(x) = lim x->0 g(x) = 0 is indeterminate. It's value is f'(0)/g'(0) assuming that both f'(0) and g'(0) aren't also both 0. We say 0/0 is an indeterminate form. >The first case in your pic is the same as above if you use h(x) = 1/g(x) and look at lim x->0 f(x) h(x). >For the second case consider f(n)^g(n) where f(n) = 1+a/n, g(n)=n. >lim n->infty f(n) = 1 >lim n->infty g(n) = inf >lim n->infty f(n)^g(n) = e^a >So it varies depending on what a you select. Why was this deleted?

Bentley Anderson

Dunno, but for legibility kindly use TeX if making any non-trivial formal math arguments.

Cooper Lopez

No, not really. It isn't any specific number, therefore [math]\infty - \infty[/math] is indeterminate, because \infty is not guaranteed to equal itself.

[math]\infty[/math] can only make any algebraic sense if it's treated as "potential infinity", but not if someone uses it to denote "actual infinity".

Chase Ward

If [math]\infty \cdot 0 := \lim_{x \to \infty} x \cdot \frac{1}{x}[/math] then math]\infty \cdot 0 = 1[/math]

Caleb Parker

This is the essence of it. Indeterminate forms do not mean what OP think they mean.

Eli Harris

All related problems seem to be stemming from the fact, that the symbol [math]\infty[/math] is being whimsically used to interchangably denote both "potential infinity" and "actual infinity". While related, they are different and cannot be arbitrarily interchanged without causing problems.

Blake Parker

f(x) = 0 g(x) = ∞ lim x -> c f(x)/ (1/g(x)) = 0/0 lim x -> c g(x) / (1 / f(x) ) = ∞/∞

QED

Daniel Evans

because your mom you fucking degenerate

David Cook

>undefined nobody said "undefined", retard

Gabriel Turner

Because it's not actually a zero, and it's not actually infinity. The product of a sequence converging to zero and a sequence diverging toward infinity does not always converge to zero.

Landon Foster

just make up an imaginary number that satisfies that condition

Jordan Wood

But you can't. [math]0 \cdot x[/math] means "none of [math]x[/math]" for any [math]x[/math] you could possibly imagine. "None of " is always just nothing, i.e. [math]0[/math].

Daniel Baker

Ok, so you mean to say that the expressions brought up by OP, if they happen to pop up in the context of limits, don't literally mean what they appear to mean algebraically?

Levi Morris

>g(x) = ∞

Camden Wilson

Because it's not real.

Charles Jones

Infinity isn't a thing, not by the definition of "thing" you're using.

Samuel Russell

2 * infinity = infinity, infinity >> 0 therefore 2 is the multiplicative identity and 2^inf = 2 prove me wrong

Jaxson Collins

0 (exact amount) multiplied by anything is NOT indeterminate.

1 (exact amount) raised to anything is NOT indeterminate.

Something tending to 0 (e.g. the limit of x as x approaches 0) multiplied by something tending to infinity is an indeterminate form.

Something tending to 1 raised to something tending to infinity is an indeterminate form.

Indeterminate forms do not mean the actual numbers 0, 1, or infinity (lol?) multiplied by or raised to one another.

Indeterminate forms only exist within the context of limits. They refer to limits of something that is the product/quotient/exponentiation of things that tend to 0, 1, or infinity.

Cooper Green

So what you're trying to say is that while 's equations are correct, the more general limits

[math]\displaystyle \lim_{x \to 0, y \to \infty}{(x \cdot y)}[/math] [math]\displaystyle \lim_{x \to 1, y \to \infty}{x^{y}}[/math]

are undefined. Is this correct?

Adrian Brooks

Infinity is not a number. If you are working with infinity, but default you are using limits. If the result for your limit is 0*infinity, you've solved it wrong. 1^infinity, 0/0, infinity*0, these aren't actual things, they're forms, they give you guidelines on how to find a limit.

Consider this: the order of magnitude we're working with is 1. Then, we could say that numbers whose absolute value is smaller than 10^-9 (and this is a very strict bound, on most any problem you don't have to go that low) is basically 0. Likewise, any number over 10^9 is too large, basically infinity. This is to scale it to less abstract ideas. We can assume these, because adding 10^-9 to any number (that makes sense to work with in our scale) basically doesn't change that number and adding any likewise number to 10^9 doesn't change 10^9. Then a*10^9 is also "infinity". Obviously, if a is too small, then it wouldn't be "infinity", but with actual infinities this is a non-issue. Lastly a*10^9 * 10^-9 is "infinity" *"zero" and is exactly a. And that a can be any number (because of our premise, it can't be much lower that 10^-5). Since there is no way for us to know what a is, infinity*0 is indeterminate.

Ethan Carter

by default*

Brody Thompson

you say >infinity is not a number, by default you're working with limits

but what you mean is >i've never seen this symbol used in any other way than when getting shit out of plug and chug in my babby calculus class

Jose Reyes

I meant what I said. Infinity is not a number.

Sebastian Diaz

that's not the point. the point is you've only seen it used in calculus and that's not its only use.

your metaphor of "infinity is just a big fixed finite number" is also useless and misguided

Bentley Kelly

I said infinity is not a number, much less a big fixed one. It's an attempt to make the problem less abstract by using the method one would use in real problems. Read with your brain, not just your eyes.

Parker Scott

there's the answer

Isaiah Murphy

While the symbol can have other meanings, none of them are relevant to elementary understanding of indeterminate forms. Would you like to argue about all of the different operations a pair of adjacent symbols can represent? The list gets bigger by the year. For a pedant, you're being stupid.

Jeremiah Morales

How the fuck did you take the exact opposite of what he said out of what he said?

Hudson Diaz

If you squeeze a rock properly, it will cause a chemical reaction where water is a product. Likewise, if you add 0 ad infinitum properly, the result will be 0. But 0 =1-1, and if you add that ad infinitum, you will not get 0. Thus, 0, addition, equality, and infinity must be defined more rigorously for 0+0+...=0.

Alexander Brown

> Check out these symbols which i've only seen in calculus > why are they indeterminate in the context of calculus when outside of this context nobody has ever seen them? > they could mean so much more than they do, like they could all mean exclusively their trivial case! 10/10 very clever.

Angel Parker

Yes. Although the term is indeterminate, not undefined.

Ian Cooper

>1-1, and if you add that ad infinitum, you will not get 0 Please don't tell me you believe that [math]\displaystyle \sum_{x=1}^{\infty}{(1-1)} = \frac{1}{2}[/math]. Please. That'd be pretty much the same tier as unironically believing that [math]\displaystyle \sum_{x=1}^{\infty}{x} = -\frac{1}{12}[/math].

Colton Gonzalez

>believing It's provable or not, in this or that theory.

Joseph Nguyen

also, 1+2+4+8+16+...=-1 what do you mean the series has to converge? why did i fail calc II?

Jayden Peterson

The limit of 1/f(x) as x approaches infinity of any unbounded and monotone increasing (past some value) is zero, you dumb dumb. x and x^2 match this description.

Also, OP, infinity isn't a real number so don't bother with real number multiplication. infinity times 0 might mean the sum from n=1 to infinity of 0, but 0=(1-1) so we can turn that sum into "the sum from n=0 to infinity of 1 minus the sum from k =0 to infinity of -1" Apply the riemann conditional convergence theorem and you'll see that this can take on any extended real value.

Also, before "muh latex," fuck off. You don't deserve it! ;^)

Hudson Martin

>2 is the multiplicative identity Goddamn. You can't just throw around words. On which algebraic structure and underlying set are you saying it's the multiplicative identity? A multiplicative identity is neutral for all elements of the set, not just for one, but 2*1=2, so 2 is non neutral on the reals.

Zachary Cruz

"Something tending towards..." Isn't a real concept, kiddo. Define the following functions on the extended reals: for all x in the E. R., f(x)=0 and g(x)=infinity. Now what is (f*g)(x)? The issue is not in the distinction between limit value and point value. The issue is in that multiplication is not well defined on the extended reals.

Sebastian Campbell

Infinity is a real number though. It's an extended real. :^(

Parker Edwards

It is any real value since it is a conditionally convergent series.

Jace Harris

>believe We define methods for manipulating infinite series and work with the consistent results that arise. In the realm of Calc 2, we are interested only in convergent series. In the rest of the world, when a divergent series appears, we must do SOMETHING with it to make progress.

No representation is sacred, we find analytic continuation more powerful for divergent series than treating divergent series as a literal infinite sum which gets us nowhere.