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k i A Letter to MSM
Arr0w   23
N Sep 19, 2022 by scannose
Greetings.

I have seen many posts talking about commonly asked questions, such as finding the value of $0^0$, $\frac{1}{0}$,$\frac{0}{0}$, $\frac{\infty}{\infty}$, why $0.999...=1$ or even expressions of those terms combined as if that would make them defined. I have made this post to answer these questions once and for all, and I politely ask everyone to link this post to threads that are talking about this issue.
[list]
[*]Firstly, the case of $0^0$. It is usually regarded that $0^0=1$, not because this works numerically but because it is convenient to define it this way. You will see the convenience of defining other undefined things later on in this post.

[*]What about $\frac{\infty}{\infty}$? The issue here is that $\infty$ isn't even rigorously defined in this expression. What exactly do we mean by $\infty$? Unless the example in question is put in context in a formal manner, then we say that $\frac{\infty}{\infty}$ is meaningless.

[*]What about $\frac{1}{0}$? Suppose that $x=\frac{1}{0}$. Then we would have $x\cdot 0=0=1$, absurd. A more rigorous treatment of the idea is that $\lim_{x\to0}\frac{1}{x}$ does not exist in the first place, although you will see why in a calculus course. So the point is that $\frac{1}{0}$ is undefined.

[*]What about if $0.99999...=1$? An article from brilliant has a good explanation. Alternatively, you can just use a geometric series. Notice that
\begin{align*}
\sum_{n=1}^{\infty} \frac{9}{10^n}&=9\sum_{n=1}^{\infty}\frac{1}{10^n}=9\sum_{n=1}^{\infty}\biggr(\frac{1}{10}\biggr)^n=9\biggr(\frac{\frac{1}{10}}{1-\frac{1}{10}}\biggr)=9\biggr(\frac{\frac{1}{10}}{\frac{9}{10}}\biggr)=9\biggr(\frac{1}{9}\biggr)=\boxed{1}
\end{align*}
[*]What about $\frac{0}{0}$? Usually this is considered to be an indeterminate form, but I would also wager that this is also undefined.
[/list]
Hopefully all of these issues and their corollaries are finally put to rest. Cheers.

2nd EDIT (6/14/22): Since I originally posted this, it has since blown up so I will try to add additional information per the request of users in the thread below.

INDETERMINATE VS UNDEFINED

What makes something indeterminate? As you can see above, there are many things that are indeterminate. While definitions might vary slightly, it is the consensus that the following definition holds: A mathematical expression is be said to be indeterminate if it is not definitively or precisely determined. So how does this make, say, something like $0/0$ indeterminate? In analysis (the theory behind calculus and beyond), limits involving an algebraic combination of functions in an independent variable may often be evaluated by replacing these functions by their limits. However, if the expression obtained after this substitution does not provide sufficient information to determine the original limit, then the expression is called an indeterminate form. For example, we could say that $0/0$ is an indeterminate form.

But we need to more specific, this is still ambiguous. An indeterminate form is a mathematical expression involving at most two of $0$, $1$ or $\infty$, obtained by applying the algebraic limit theorem (a theorem in analysis, look this up for details) in the process of attempting to determine a limit, which fails to restrict that limit to one specific value or infinity, and thus does not determine the limit being calculated. This is why it is called indeterminate. Some examples of indeterminate forms are
\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]etc etc. So what makes something undefined? In the broader scope, something being undefined refers to an expression which is not assigned an interpretation or a value. A function is said to be undefined for points outside its domain. For example, the function $f:\mathbb{R}^{+}\cup\{0\}\rightarrow\mathbb{R}$ given by the mapping $x\mapsto \sqrt{x}$ is undefined for $x<0$. On the other hand, $1/0$ is undefined because dividing by $0$ is not defined in arithmetic by definition. In other words, something is undefined when it is not defined in some mathematical context.

WHEN THE WATERS GET MUDDIED

So with this notion of indeterminate and undefined, things get convoluted. First of all, just because something is indeterminate does not mean it is not undefined. For example $0/0$ is considered both indeterminate and undefined (but in the context of a limit then it is considered in indeterminate form). Additionally, this notion of something being undefined also means that we can define it in some way. To rephrase, this means that technically, we can make something that is undefined to something that is defined as long as we define it. I'll show you what I mean.

One example of making something undefined into something defined is the extended real number line, which we define as
\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]So instead of treating infinity as an idea, we define infinity (positively and negatively, mind you) as actual numbers in the reals. The advantage of doing this is for two reasons. The first is because we can turn this thing into a totally ordered set. Specifically, we can let $-\infty\le a\le \infty$ for each $a\in\overline{\mathbb{R}}$ which means that via this order topology each subset has an infimum and supremum and $\overline{\mathbb{R}}$ is therefore compact. While this is nice from an analytic standpoint, extending the reals in this way can allow for interesting arithmetic! In $\overline{\mathbb{R}}$ it is perfectly OK to say that,
\begin{align*}
a + \infty = \infty + a & = \infty, & a & \neq -\infty \\
a - \infty = -\infty + a & = -\infty, & a & \neq \infty \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \pm\infty, & a & \in (0, +\infty] \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \mp\infty, & a & \in [-\infty, 0) \\
\frac{a}{\pm\infty} & = 0, & a & \in \mathbb{R} \\
\frac{\pm\infty}{a} & = \pm\infty, & a & \in (0, +\infty) \\
\frac{\pm\infty}{a} & = \mp\infty, & a & \in (-\infty, 0).
\end{align*}So addition, multiplication, and division are all defined nicely. However, notice that we have some indeterminate forms here which are also undefined,
\[\infty-\infty,\frac{\pm\infty}{\pm\infty},\frac{\pm\infty}{0},0\cdot \pm\infty.\]So while we define certain things, we also left others undefined/indeterminate in the process! However, in the context of measure theory it is common to define $\infty \times 0=0$ as greenturtle3141 noted below. I encourage to reread what he wrote, it's great stuff! As you may notice, though, dividing by $0$ is undefined still! Is there a place where it isn't? Kind of. To do this, we can extend the complex numbers! More formally, we can define this extension as
\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]which we call the Riemann Sphere (it actually forms a sphere, pretty cool right?). As a note, $\tilde{\infty}$ means complex infinity, since we are in the complex plane now. Here's the catch: division by $0$ is allowed here! In fact, we have
\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]where $\tilde{\infty}/\tilde{\infty}$ and $0/0$ are left undefined. We also have
\begin{align*}
z+\tilde{\infty}=\tilde{\infty}, \forall z\ne -\infty\\
z\times \tilde{\infty}=\tilde{\infty}, \forall z\ne 0
\end{align*}Furthermore, we actually have some nice properties with multiplication that we didn't have before. In $\mathbb{C}^*$ it holds that
\[\tilde{\infty}\times \tilde{\infty}=\tilde{\infty}\]but $\tilde{\infty}-\tilde{\infty}$ and $0\times \tilde{\infty}$ are left as undefined (unless there is an explicit need to change that somehow). One could define the projectively extended reals as we did with $\mathbb{C}^*$, by defining them as
\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]They behave in a similar way to the Riemann Sphere, with division by $0$ also being allowed with the same indeterminate forms (in addition to some other ones).
23 replies
Arr0w
Feb 11, 2022
scannose
Sep 19, 2022
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A Letter to MSM
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Arr0w
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Greetings.

I have seen many posts talking about commonly asked questions, such as finding the value of $0^0$, $\frac{1}{0}$,$\frac{0}{0}$, $\frac{\infty}{\infty}$, why $0.999...=1$ or even expressions of those terms combined as if that would make them defined. I have made this post to answer these questions once and for all, and I politely ask everyone to link this post to threads that are talking about this issue.
  • Firstly, the case of $0^0$. It is usually regarded that $0^0=1$, not because this works numerically but because it is convenient to define it this way. You will see the convenience of defining other undefined things later on in this post.
  • What about $\frac{\infty}{\infty}$? The issue here is that $\infty$ isn't even rigorously defined in this expression. What exactly do we mean by $\infty$? Unless the example in question is put in context in a formal manner, then we say that $\frac{\infty}{\infty}$ is meaningless.
  • What about $\frac{1}{0}$? Suppose that $x=\frac{1}{0}$. Then we would have $x\cdot 0=0=1$, absurd. A more rigorous treatment of the idea is that $\lim_{x\to0}\frac{1}{x}$ does not exist in the first place, although you will see why in a calculus course. So the point is that $\frac{1}{0}$ is undefined.
  • What about if $0.99999...=1$? An article from brilliant has a good explanation. Alternatively, you can just use a geometric series. Notice that
    \begin{align*}
\sum_{n=1}^{\infty} \frac{9}{10^n}&=9\sum_{n=1}^{\infty}\frac{1}{10^n}=9\sum_{n=1}^{\infty}\biggr(\frac{1}{10}\biggr)^n=9\biggr(\frac{\frac{1}{10}}{1-\frac{1}{10}}\biggr)=9\biggr(\frac{\frac{1}{10}}{\frac{9}{10}}\biggr)=9\biggr(\frac{1}{9}\biggr)=\boxed{1}
\end{align*}
  • What about $\frac{0}{0}$? Usually this is considered to be an indeterminate form, but I would also wager that this is also undefined.
Hopefully all of these issues and their corollaries are finally put to rest. Cheers.

2nd EDIT (6/14/22): Since I originally posted this, it has since blown up so I will try to add additional information per the request of users in the thread below.

INDETERMINATE VS UNDEFINED

What makes something indeterminate? As you can see above, there are many things that are indeterminate. While definitions might vary slightly, it is the consensus that the following definition holds: A mathematical expression is be said to be indeterminate if it is not definitively or precisely determined. So how does this make, say, something like $0/0$ indeterminate? In analysis (the theory behind calculus and beyond), limits involving an algebraic combination of functions in an independent variable may often be evaluated by replacing these functions by their limits. However, if the expression obtained after this substitution does not provide sufficient information to determine the original limit, then the expression is called an indeterminate form. For example, we could say that $0/0$ is an indeterminate form.

But we need to more specific, this is still ambiguous. An indeterminate form is a mathematical expression involving at most two of $0$, $1$ or $\infty$, obtained by applying the algebraic limit theorem (a theorem in analysis, look this up for details) in the process of attempting to determine a limit, which fails to restrict that limit to one specific value or infinity, and thus does not determine the limit being calculated. This is why it is called indeterminate. Some examples of indeterminate forms are
\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]etc etc. So what makes something undefined? In the broader scope, something being undefined refers to an expression which is not assigned an interpretation or a value. A function is said to be undefined for points outside its domain. For example, the function $f:\mathbb{R}^{+}\cup\{0\}\rightarrow\mathbb{R}$ given by the mapping $x\mapsto \sqrt{x}$ is undefined for $x<0$. On the other hand, $1/0$ is undefined because dividing by $0$ is not defined in arithmetic by definition. In other words, something is undefined when it is not defined in some mathematical context.

WHEN THE WATERS GET MUDDIED

So with this notion of indeterminate and undefined, things get convoluted. First of all, just because something is indeterminate does not mean it is not undefined. For example $0/0$ is considered both indeterminate and undefined (but in the context of a limit then it is considered in indeterminate form). Additionally, this notion of something being undefined also means that we can define it in some way. To rephrase, this means that technically, we can make something that is undefined to something that is defined as long as we define it. I'll show you what I mean.

One example of making something undefined into something defined is the extended real number line, which we define as
\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]So instead of treating infinity as an idea, we define infinity (positively and negatively, mind you) as actual numbers in the reals. The advantage of doing this is for two reasons. The first is because we can turn this thing into a totally ordered set. Specifically, we can let $-\infty\le a\le \infty$ for each $a\in\overline{\mathbb{R}}$ which means that via this order topology each subset has an infimum and supremum and $\overline{\mathbb{R}}$ is therefore compact. While this is nice from an analytic standpoint, extending the reals in this way can allow for interesting arithmetic! In $\overline{\mathbb{R}}$ it is perfectly OK to say that,
\begin{align*}
a + \infty = \infty + a & = \infty, & a & \neq -\infty \\
a - \infty = -\infty + a & = -\infty, & a & \neq \infty \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \pm\infty, & a & \in (0, +\infty] \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \mp\infty, & a & \in [-\infty, 0) \\
\frac{a}{\pm\infty} & = 0, & a & \in \mathbb{R} \\
\frac{\pm\infty}{a} & = \pm\infty, & a & \in (0, +\infty) \\
\frac{\pm\infty}{a} & = \mp\infty, & a & \in (-\infty, 0).
\end{align*}So addition, multiplication, and division are all defined nicely. However, notice that we have some indeterminate forms here which are also undefined,
\[\infty-\infty,\frac{\pm\infty}{\pm\infty},\frac{\pm\infty}{0},0\cdot \pm\infty.\]So while we define certain things, we also left others undefined/indeterminate in the process! However, in the context of measure theory it is common to define $\infty \times 0=0$ as greenturtle3141 noted below. I encourage to reread what he wrote, it's great stuff! As you may notice, though, dividing by $0$ is undefined still! Is there a place where it isn't? Kind of. To do this, we can extend the complex numbers! More formally, we can define this extension as
\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]which we call the Riemann Sphere (it actually forms a sphere, pretty cool right?). As a note, $\tilde{\infty}$ means complex infinity, since we are in the complex plane now. Here's the catch: division by $0$ is allowed here! In fact, we have
\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]where $\tilde{\infty}/\tilde{\infty}$ and $0/0$ are left undefined. We also have
\begin{align*}
z+\tilde{\infty}=\tilde{\infty}, \forall z\ne -\infty\\
z\times \tilde{\infty}=\tilde{\infty}, \forall z\ne 0
\end{align*}Furthermore, we actually have some nice properties with multiplication that we didn't have before. In $\mathbb{C}^*$ it holds that
\[\tilde{\infty}\times \tilde{\infty}=\tilde{\infty}\]but $\tilde{\infty}-\tilde{\infty}$ and $0\times \tilde{\infty}$ are left as undefined (unless there is an explicit need to change that somehow). One could define the projectively extended reals as we did with $\mathbb{C}^*$, by defining them as
\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]They behave in a similar way to the Riemann Sphere, with division by $0$ also being allowed with the same indeterminate forms (in addition to some other ones).
This post has been edited 13 times. Last edited by Arr0w, Sep 17, 2022, 11:43 PM
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greenturtle3141
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#12 • 47 Y
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My several cents:
  • I think we should all just accept $0^0 = 1$ and move on with it, really no reason not to (the rationale being, "Because it is").
  • $\infty$ does not necessarily refer to the limit of a function. Honestly just reject any "arithmetic" with $\infty$ unless you have properly defined it. So we should not bother computing even "obvious" quantities like $\infty+\infty$ or $\infty + 7$. If you don't define what $\infty$ is then there is no point doing anything with it. As a side effect you get un-definedness of $\infty/\infty$ etc. for free, because... we literally have not defined $\infty$. Tada.
  • To wit, I will complain that $\infty$ should not necessarily represent the limit of a function. But if you are interpreting it as such, then $\infty/\infty$ is not undefined, but rather indeterminate.
  • At some point in math (which is not anywhere within 5 years if you're in middle school), we do start messing with $+\infty$ as a perfectly valid number, because it starts becoming useful. Particularly, we do define $0 \cdot \infty = 0$ in contexts such as measure theory. This differs from the "limit of a function" interpretation, in which $0 \cdot \infty$ would be indeterminate.

Digression
This post has been edited 2 times. Last edited by greenturtle3141, Feb 12, 2022, 4:05 AM
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Kempu33334
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#17 • 13 Y
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I think it would be good for the 0/0 one to be said as indeterminate
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Arr0w
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#26 • 9 Y
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greenturtle3141 wrote:
At some point in math (which is not anywhere within 5 years if you're in middle school), we do start messing with $+\infty$ as a perfectly valid number, because it starts becoming useful. Particularly, we do define $0 \cdot \infty = 0$ in contexts such as measure theory.
You seem to mention this quite a bit in your past posts as well. Could you clarify why this would be used and is necessary in measure theory? I think it would be a cool addition to the original post. Thanks!
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greenturtle3141
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#39 • 18 Y
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Arr0w wrote:
greenturtle3141 wrote:
At some point in math (which is not anywhere within 5 years if you're in middle school), we do start messing with $+\infty$ as a perfectly valid number, because it starts becoming useful. Particularly, we do define $0 \cdot \infty = 0$ in contexts such as measure theory.
You seem to mention this quite a bit in your past posts as well. Could you clarify why this would be used and is necessary in measure theory? I think it would be a cool addition to the original post. Thanks!

Measure Zero Differences

When you change very few points of a function, its integral should not change. For example, we know that $\int_0^1 x^2\,dx = \frac13$. But what about e.g. $\int_0^1 f(x)\,dx$ where $f(x) := \begin{cases}x^2, & x \ne \frac12 \\ 7, & x = \frac12\end{cases}$? I have changed a single point. Of course, this shouldn't matter because a single point is negligible for changing the area under the curve. If you think about it, I'm essentially adding in a rectangle with dimensions $0 \times 7$, which is zero. So $\int_0^1 f(x)\,dx = \frac13$.

In general, any measure zero difference cannot change an integral.

Infinity

Let $\overline{\mathbb{R}} := \mathbb{R} \cup \{\pm \infty\}$. This is the extended reals, and is a useful abstraction.

One way in which this makes mathematicians happy is in making limits exist. For example, consider the monotone convergence theorem, which states that if a sequence of functions $f_n:E \to \overline{\mathbb{R}}$ is increasing (i.e. $f_n \leq f_{n+1}$ everywhere in $E$, for all $n$), then $f_n$ converges pointwise to a function $f:E \to \overline{\mathbb{R}}$ and moreover
$$\lim_{n \to \infty} \int_E f_n\,dx = \int_E f\,dx.$$Notice how I allow my functions to take values of $\infty$. If I didn't let them do that, then I have to take separate cases in the statement of the theorem as to whether limits diverge or whatnot. But as you can see, there really is no issue if I let functions take values of $\infty$.

This also implies that I can integrate functions that take values of $\infty$... indeed I certainly can. What happens?

Why $\infty \times 0$ shows up

Ok, let's define this function:
$$f(x) := \begin{cases}x^2, & x \ne \frac12 \\ +\infty, x = \frac12\end{cases}$$What is $\int_0^1 f(x)\,dx$?

In the definition of Lebesgue integration, you'll find the possibility that you have to consider the "rectangle" $\{1/2\} \times [0,\infty)$. The dimensions of this are $0 \times \infty$. Well? Does this change the integral?

Here's the thing: If it did, it would be really stupid and annoying. Remember, measure-zero differences should NOT do anything to an integral. So we force $0 \times \infty := 0$ to make this work, so that $\int_0^1 f(x)\,dx = 1/3$ still.

This isn't contrived, this makes sense. A $0 \times \infty$ rectangle has no area. It doesn't cover any significant 2D space. So its area is zero. If you're pedantic, you can even prove it using the rigorous definition of area. In any case, this is a context in which it is absolutely, 100% clear what the value of $0 \times \infty$ should be. It is zero. No other value for it would be useful, no other value would make remotely any sense. This is the only possible value for it here.
This post has been edited 1 time. Last edited by greenturtle3141, Feb 23, 2022, 5:23 AM
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Nickelslordm
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@bove but infinity has no concrete definition; therefore, how can we even tell that it follows rules such as multiplication by zero? I don't know for sure, but I would guess that infinity bends the whole concept of multiplication. Someone whose name I forgot said that zero times infinity equals infinity divided by 2.
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Facejo
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$\frac{0}{0}$ is indeterminate not undefined
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Arr0w
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#46 • 6 Y
Y by HWenslawski, Kea13, ImSh95, WiseTigerJ1, TeamFoster-Keefe4Ever, chriscassano
Facejo wrote:
$\frac{0}{0}$ is indeterminate not undefined
Thank you, this has been revised.
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Facejo
2779 posts
#47 • 7 Y
Y by HWenslawski, jmiao, Kea13, ImSh95, addyc, WiseTigerJ1, TeamFoster-Keefe4Ever
Facejo wrote:
$\frac{0}{0}$ is indeterminate not undefined

wait what's the difference

Indeterminate means there is no specific value, while undefined means it doesn't exist

@above You're welcome
This post has been edited 1 time. Last edited by Facejo, Apr 14, 2022, 11:58 PM
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ssbgm9002
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#48 • 5 Y
Y by HWenslawski, Kea13, ImSh95, WiseTigerJ1, TeamFoster-Keefe4Ever
jmiao wrote:
Facejo wrote:
$\frac{0}{0}$ is indeterminate not undefined

wait what's the difference
Great explanation here
"The big difference between undefined and indeterminate is the relationship between zero and infinity. When something is undefined, this means that there are no solutions. However, when something in(is GET YOUR GRAMMAR RIGHT) indeterminate, this means that there are infinitely many solutions to the question." - http://5010.mathed.usu.edu/Fall2018/LPierson/indeterminateandundefined.html#:~:text=The%20big%20difference%20between%20undefined,many%20solutions%20to%20the%20question.
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NightFury101
1096 posts
#61 • 6 Y
Y by HWenslawski, Turtle09, Kea13, ImSh95, WiseTigerJ1, TeamFoster-Keefe4Ever
Just a picky note on the $\frac10$ thing:

If we have $f(x) = \frac1x$, then $\lim_{x \to 0^+}f(x) = \infty$ and $\lim_{x \to 0^-}f(x) = -\infty$. In other words, $f$ approaches positive infinity as $x$ approaches $0$ from the right hand side, and $f$ approaches negative infinity as $x$ approaches $0$ from the left hand side.

This is why the limit does not exist and $\frac10$ is undefined.
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YaoAOPS
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#66 • 5 Y
Y by Kea13, ImSh95, WiseTigerJ1, TeamFoster-Keefe4Ever, chriscassano
Arr0w wrote:
Greetings.

I have seen many posts talking about commonly asked questions, such as finding the value of $0^0$, $\frac{1}{0}$,$\frac{0}{0}$, $\frac{\infty}{\infty}$, why $0.999...=1$ or even expressions of those terms combined as if that would make them defined. I have made this post to answer these questions once and for all, and I politely ask everyone to link this post to threads that are talking about this issue.

It'd be nice to have the other indeterminate forms mentioned, i.e. $\infty - \infty, 1^{\infty}$. Very cool post
This post has been edited 3 times. Last edited by YaoAOPS, May 30, 2022, 3:15 PM
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Turtle09
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#68 • 5 Y
Y by Kea13, ImSh95, WiseTigerJ1, TeamFoster-Keefe4Ever, mithu542
YaoAOPS wrote:
Arr0w wrote:
Greetings.

I have seen many posts talking about commonly asked questions, such as finding the value of $0^0$, $\frac{1}{0}$,$\frac{0}{0}$, $\frac{\infty}{\infty}$, why $0.999...=1$ or even expressions of those terms combined as if that would make them defined. I have made this post to answer these questions once and for all, and I politely ask everyone to link this post to threads that are talking about this issue.

It'd be nice to have the other indeterminate forms mentioned, i.e. $\infty - \infty, 1^{\infty}$. Very cool post

wait isn't $1^\infty = 1$?
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Facejo
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#69 • 3 Y
Y by Kea13, ImSh95, WiseTigerJ1
@above No. In general, you cannot say that.
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aayr
2421 posts
#70 • 5 Y
Y by mighty_champ, Kea13, ImSh95, WiseTigerJ1, Kawhi2
wait why not
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mahaler
3075 posts
#71 • 3 Y
Y by Kea13, ImSh95, WiseTigerJ1
Turtle09 wrote:
wait isn't $1^\infty = 1$?
Facejo wrote:
@above No. In general, you cannot say that.
aayr wrote:
wait why not

Yeah, I don't know either...

Couldn't you just prove that $1^n = 1$ for any positive integer $n$ using induction?
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Lionking212
5070 posts
#72 • 3 Y
Y by Kea13, ImSh95, WiseTigerJ1
mahaler wrote:
Turtle09 wrote:
wait isn't $1^\infty = 1$?
Facejo wrote:
@above No. In general, you cannot say that.
aayr wrote:
wait why not

Yeah, I don't know either...

Couldn't you just prove that $1^n = 1$ for any positive integer $n$ using induction?

because
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Facejo
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#81 • 4 Y
Y by Kea13, ImSh95, WiseTigerJ1, Elephant200
Post #70 by aayr

Post #71 by mahaler

Consider the limit $\displaystyle \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\approx 2.718281828459045$
This post has been edited 1 time. Last edited by Facejo, May 30, 2022, 7:06 PM
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Arr0w
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#96 • 3 Y
Y by Kea13, ImSh95, WiseTigerJ1
RaymondZhu wrote:
Can you please define indeterminate and undefined and the differences in your post? Thanks!
Howdy Raymond!

I have made sure to add some additional items like you requested. If there's anything more you guys want to see from this thread let me know so I can add/change it. Thanks!
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Arr0w
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#130 • 3 Y
Y by Kea13, ImSh95, WiseTigerJ1
peelybonehead wrote:
Leo2020 wrote:
bump$            $
We should bump every single day so people can be reminded of this thread every single day
Please don't do this. If you would like to share this post for whatever reason, you can just link it using the following format:
Please see [url=https://artofproblemsolving.com/community/c3h2778686_a_letter_to_msm] here [/url].

In the meantime, I have made some additional edits to the letter as I have changed my mind on some conclusions I made previously. If there are any discrepancies, please let me know. Thank you.
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wamofan
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#140 • 3 Y
Y by Kea13, ImSh95, WiseTigerJ1
thebluepenguin21 wrote:
but only 1=1, right? So you can't say that 0.99999999... = 1.
that's like saying only 2=2, so you can't say that 6-4=2; completely wrong
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Sure it makes sense because that is the closest that we can get, but it can not be. And 9x -x = 8.99999999999... So this can not be true.

why is 9x-x=8.99999?
9x=9, x=1 so 8x=8 so x=1
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ap246
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#146 • 5 Y
Y by Kea13, ImSh95, s12d34, CatsMeow12, WiseTigerJ1
Basically, we can just use the Epsilon - Delta definition of a limit. If you give me an $\epsilon > 0$ such that $\epsilon = 1 - 0.99999\dots$ then I will provide a sufficient $\delta$ for the number of nines needed such that $1 - 0.99999\dots < \epsilon$ We can use this to form a proof by contradiction. More specifically, for a positive value $x,$ if $\epsilon > 10^{-x}$ then $\delta = x$ is sufficient. Therefore, there will always be a delta such that $1 - 0.99999\dots < \epsilon$

We can define the limit: $$\lim_{x\to \infty} f(x) = y$$if for every $\epsilon > 0$ there exists a $\delta$ such that $x > \delta$ which implies $$|f(x) - y| < \epsilon$$so $f(x) = \sum_{n = 1}^{n = x} 9 * 10^{-n}$ for a positive value of $n.$

If $2$ quantities aren't equal, then there must be a nonzero difference, but given that for every $\epsilon$ there exists a $\delta$ such that the given information is met, we prove that there doesn't exist a nonzero difference, so both quantities are equal.

$$Q.E.D$$
This post has been edited 2 times. Last edited by ap246, Sep 19, 2022, 8:03 PM
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scannose
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#147 • 4 Y
Y by Kea13, ImSh95, megahertz13, MathematicalGymnast573
Also, if you agree that geometric series works:
$0.999\dots=9({1\over{10}}+{1\over{100}}+\dots)=9\cdot{1\over{1-{1\over{10}}}}=9\cdot{1\over9}=1$
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scannose
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#149 • 9 Y
Y by mathfan2020, Kea13, lucaswujc, ImSh95, MathematicalGymnast573, megahertz13, WiseTigerJ1, TeamFoster-Keefe4Ever, Blue_banana4
I'm pretty sure that 9x-x=8.999... was just a typo; they probably meant 10x-x=8.999...?
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