# The Curious Case of Nothing multiplied No times

Recently, a question that I posted raised a healthy debate.

The question was to find $\lim_{x \to 0} x^x$.

There were various veiwpoints, none of them illogical.

Viewpoint 1 : It is complete gobblydook

The given problem can be written as :

$\lim_{x \to 0} x^x = \lim_{a \to 0} \lim_{b \to 0} a^b$

$=\lim_{a \to 0} a^0 = \lim_{a \to 0} 1 = 1$

However, exchanging the limits,

$\lim_{x \to 0} x^x = \lim_{b \to 0} \lim_{a \to 0} a^b$

$=\lim_{b \to 0} 0^b = 0, \pm \infty$ depending on $b \to 0^+$ or $b \to 0^-$

Hence, even if we consider only the left OR right limit, there is ambiguity.

Hence, it is gobblydook.

Viewpoint 2 : The Left is not Right

No, No... You are not to make a mess of the problem by taking two 'limits', it is the same quantity that is varying.

The right limit is (Courtesy : Ryan Tamburrino)

$\lim_{x \to 0^+} x^x = \exp\left(\lim_{x \to 0^+}x\ln(x)\right)=\exp\left(\lim_{x \to 0^+}\frac{\ln(x)}{1/x}\right)$

Applying L'Hopital's rule,

$\lim_{x \to 0^+} x^x = \exp\left(\lim_{x \to 0^+}\frac{1/x}{-1/x^2}\right)=\exp(\lim_{x \to 0^+} -x)=1$

The left limit involves complex exponentiation, fractional power of negative numbers and all other mathematical boggarts. It would be a bad idea to go into that lest we fall prey to tetration.

Viewpoint 3 : All's well - The limits agree

The right limit is indeed 1, as shown in Viewpoint 2. However, there is no need to panic as the left limit is also a tame one, and certainly not a boggart.

For the left limit, consider $x=-r$ with $r$ being positive. The problem now becomes

$\lim_{r \to 0^+} (-r)^{-r} = \exp\left(\lim_{r \to 0^+}-r\left(\ln(r)+\ln(-1)\right)\right)$

$= \exp\left(\lim_{r \to 0^+}-r(\ln(r)+j\pi)\right) = \exp\left(-\lim_{r \to 0^+}r\ln(r)-\lim_{r \to 0}jr\pi\right)$

$=\exp\left(0+0\right)=1$

Whom do you agree with?

What is your opinion on the 'value' of $0^0$. (Not many seem to want to fix a value for it!!).

What is your opinion on the existence/agreement/value of the limits? Note by Janardhanan Sivaramakrishnan
4 years, 11 months ago

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Thanks for posting this "curious case" as a note. Later, I'll bring up some points why it's still an unsettled matter, although for most practical purposes it's said to be "$1$" . This is a good example why it's not true that "everything" in mathematics, "if pursued with rigor" necessarily lead to unambiguous conclusions. Sometimes it does not.

- 4 years, 11 months ago

But there is a huge difference between the term $\frac{0}{0}$ and the limit $\lim _{ x \rightarrow 0 } \frac{ x}{x}$.

It is true that if $f(x) , g(x)$ are non-zero polynomials such that $f(0) = g(0) = 0$, then $\lim_{x \rightarrow 0 } f(x) ^ { g(x) } = 1$.

Staff - 4 years, 11 months ago

Calvin, I seem to recall both you and I discussing a problem where a function had a limiting value for a particular $x$, and yet at that particular value of $x$ it was indeterminate, and therefore was not a root. This is the same thing happening here again. I agree that the limit of $\displaystyle {x}^{x}$ as $x\rightarrow 0$ is $1$, but that does not mean that $0^0$ has a determinate value. Here is an easy example to "show why" it can't be determinate:

${ 0 }^{ 0 }= 0 }^{ 1-1 }=\dfrac { 0 }{ 0 }$

which we know is indeterminate (the very point you had argued for with that problem with roots!)

More generally, if we consider the function $f(x,y) = {x}^{y}$, and we examine the double limit $x\rightarrow 0$ and $y\rightarrow 0$, then in this case there isn't even a consistent limiting value. In other words, there is a "nice" limiting value ONLY for the single-valued function $f(x) = {x}^{x}$, but it's risky to assume that ${0}^{0}=1$ in any context. It all depends.

- 4 years, 11 months ago

You misunderstand me. At no point am I saying that $0 ^ 0 = 1$. I am not saying that $0 ^ 0 = \lim_{ x \rightarrow 0 } x ^ x$. In fact, quite the opposite, I am saying that there is a "huge difference" between the two of them.

What I am saying is that $1 = \lim_{ x \rightarrow 0 } x ^ x$, which was what I thought the question is about.

To me, $0^0$ it is indeterminate, though yes there are reasons why we might want to define it as $1$.

Staff - 4 years, 11 months ago

Were you responding to my first comment here, or to my reply to Jake Lai? But, yes, the question was about the limit ${x}^{x}$ as $x\rightarrow 0$ , and, yes, it's $1$. Now that we're on this matter, when I made my first comment disputing the originally posted problem, I later found that there wasn't an easy way to come back and try to clarify it without either 1) making an addendum to my original comment, or 2) as a reply to someone else. Is there a way for me to enter a new comment, in an effort to clarify? I ended up making an addendum to my original comment, but I don't know if anyone has picked that up, since it won't notify anyone that a new comment has been made.

- 4 years, 11 months ago

Ah, in which case I think I misunderstood you. Sounds like we are in agreement.

Ah, now I see your edit, and I understand it better. Right, in a sense, if you think it's worthwhile to inform everyone else in the discussion, then yes reply. If it's a minor correction / addition, then hit edit.

(Ignore this)

With regards to reports, if you hit on the Reply button, that will open up a reply box Staff - 4 years, 11 months ago

Let me go check it out and try it.

- 4 years, 11 months ago

$f(x) = x^{x}$ is well-defined everywhere except at $x = 0$ (if we're accepting complex values, of course). The two-sided limit is 1. Also note, you cannot just switch from a limit of one variable ($x$ in this case) to one of two variables ($a, b$ here).

- 4 years, 11 months ago

$f(x)=\dfrac { x }{ x }$ is also well-defined everywhere except at $x=0$, and yet $\dfrac { 0 }{ 0 }$ is indeterminate.
In both cases, the limiting value is $1$, but both $\displaystyle 0^0$ and $\dfrac { 0 }{ 0 }$ are undefined.

- 4 years, 11 months ago

0/ 0 is indeterminate, means something to me. For 0 x n = 0, dividing both sides by 0 obtained n = 0/ 0, let say.

Although 0^0 = 0^(N - N) = 0^N/ 0^N, it will not be having different N and therefore definable to be 1.

1 is one among all from indeterminate.

Limit x--> 3 for (x - 3)/ (x^2 - 9) = 1/ 6 which is not necessarily 1, but Limit x --> 3 for (x^2 - 9)^(x - 3) is 1.

Therefore, 0^0 <> 0/ 0.

A 0/ 0 can be 0^0 but 0^0 is not 0/ 0, as 0/ 0 can stand for all others which do not stand for 0^0.

0^0 simply definable to be (exact 0)^(exact 0) = 1. Just be careful when we take limits of a form looked like 0^0.

Note: Limit is a subject to mean for as thought as exact and say as exact because we cannot know how tiny the smallest thing is to be described. Just like I ask you, do you see a quark? Something even smaller than this is still valid but just like seeing nothing. This is the concept of limit I believe in.

- 4 years, 11 months ago

I think you would enjoy the short story The Indifatigable Frog. It's about an argument between a scientist and a philosopher and explores concepts of the limit, such as the barrier between the world of math and physical reality.

https://phillipkay.files.wordpress.com/2011/07/collected-stories-1-the-short-happy-life.pdf

- 4 years, 11 months ago

In truth, we don't talk about authority but reasonable reasons. The important thing is straight to the fact, and our attitude or expectation is just to sort for a true answer. Too long for me to read and I don't think I ought to do this. Direct to questioning and answering is the right thing to do for truth.

- 4 years, 11 months ago

Well, long story short, the scientist builds a machine to test a paradox. The machine has a tunnel, a frog, and a shrinking ray. Each time the frog jumps, the frog is shrunk to half of its original size, making the length of the jump also half. The scientist believes that the frog will never reach the end, and the philosopher believes that the frog will make it to the end. The scientist goes crazy and traps the philosopher in the machine, and the philosopher keeps moving and shrinking until eventually he falls through the tunnel due to spaces in matter.

When the philosopher escapes he returns to his original size, and the scientist and the philosopher implicitly agree that there may not be a direct solution to the problem applicable to both the physical world and the world of math.

I do agree that direct questioning is the best way to find truth, but authority is necessary to some extent to help us start from something instead of nothing.

- 4 years, 11 months ago

Mathematics doesn't concern itself with physical reality or even philosophy. In the context of this problem, the frog , in the limit, travels the distance of the tunnel, but it's not necessarily true that it actually makes it to end of it.

Now, a physicist, which normally isn't necessarily a rigorous mathematician, will argue that if the frog makes each jump in proportionally less time than the previous, then in the limiting case it will reach the end of the tunnel in a finite time, and therefore we can surmise he will reach the end of the tunnel for all practical purposes. A lot of theoretical physics is like that. Just as mathematics doesn't concern itself with physical reality, theoretical physics doesn't concern itself with mathematical rigor---and leaves philosophy to the philosophers.

- 4 years, 11 months ago

Authority ought to be there for any changes but we don't talk about authority just for stopping any reasonable reasons or proper reasons; this is darkness. Provided correct and reasonable, we ought to take as it is. (Please regard these sentences as out of topic, which don't really answer to question.)

1/ 2 + 1/ 4 + 1/ 8 + 1/ 16 + ... = (1/ 2)/(1 - 1/ 2) = 1. The frog ought to be able to jump many times. Assuming half the original journey for the first jump, the limit is 1. I don't think this is a contradiction between physics and mathematics though.

Could you give an example that physical phenomena that do not agree with mathematics?

- 4 years, 11 months ago

But you are speaking about general $N$; when $N$ is not a positive real, $0^{0} = 0^{N-N} = \frac{0^{N}}{0^{N}} = 1$ makes no sense. You have to move to limits to say that $\lim_{x \rightarrow 0} x^{x} = 1$; for "exact 0" it doesn't work.

Also, for $N = 1$, doesn't that contradict what you just said before? (That $\frac{0}{0}$ is indeterminate.) Moreover, we are dealing with equality - which is transitive - so what do you mean by "a $\frac{0}{0}$ can be $0^{0}$ but $0^{0}$ is not $\frac{0}{0}$"?

Also, no offense, but nobody cares about what you imagine the limit to be; the limit is a mathematical concept and doesn't concern itself with how the physical world works (just as Michael Mendrin said).

- 4 years, 11 months ago

Limit x--> 0 x^x = 1 is not of doubt. 0^(N - N) is to illustrate how it can differ from 0^M/ 0^N where M <> N, if you finish reading all my writings. "Exact 0" doesn't work, this is true. For d y/ d x, if d x is exact 0, then d y/ d x shall also be the same. By mean of defining 0^0 = 1, we do the same as a symbolized resemblance to appreciate the typical ratio of the limit x--> 0 x^x, which is assigned as 0^0. Just like d y/ d x or limit x--> 0 for sin x/ x, no one is of exact zero denominator; but we must have a symbol d y/ d x to describe and represent the fact.

When you explain d x to me, you shall find trouble of how you can express to me. As thought as exact but not actually exact while still saying exact is where we find some mean to express a limit. It is trivial to explain to people that all denominator of d y/ d x is not exactly zero. The reason why I use physics to imagine the most tiny in limit is when we are actually asked to elaborate what d x is, we shall automatically think back to what is actual.

I believe you learnt about set theory. 1 is one among all. In other words, 1 is a member of indeterminate. 0/ 0 is indeterminate while 0^0 is 1. Therefore, 0^0 is a member of 0/ 0. 0/ 0 is something which covers all while 0^0 can only be 1.

- 4 years, 11 months ago

Log x {base x} = Log x/ Log x = 1 = Log x, for x <>1 let say.

With x = x^Log x whenever we like,

x^(1/ Log x) = x as well, as [x^(Log x)]^(1/ Log x) = [x]^(1/ Log x) or just because 1/ 1 = 1 with Log x = 1.

Log x = 1 is itself an equation for indeterminate to some x.

When x^(Log x) is written with base x, Log x = 1 and x itself has been an indeterminate.

Now x^(Log x) is purposely aimed for [x^(Log x)]^(1/ Log x) = x,

Meaning to say x has escaped from being x --> 0 but some a or e.

On the other hand for [a^(Log x)]^(1/ Log x),

The limit for x --> 0 becomes a which includes e as a possible base, where x --> 0 is not being against.

However, please remember that (1/ Log x) here is having base a which is a form of indeterminate. You get what you wanted but not actually doing a proper limit with fixed value.

I think putting (1/ Log x) as a (0-) for x of (0) for x^(1/ Log x) = 0^(0-) here has introduced some base a to decide for its value of the limit. We may think of x--> 0- for (0-)^(0-) to argue for same thing for 0^0, but at least a power to a Log a/ Log x {saying explicitly} ought to be excepted because of features of logarithm itself.

Limit x --> 0 for such x^[Log a/ Log x] = a is just x^(Log a) = a without bothering about x --> 0. Antilog (Log a) = a don't bother about what x is being described. That was why one of our friend here asked if it doesn't matter of what x is. Then, the limit is not under a discussion of limit. Quoting his question: "Even without the limit, does it suggest that x^(1/ Log x) = e, for all x where the function is defined??"

- 4 years, 11 months ago

Has anyone taken into consideration the fact that a calculator might be useful ? we know that f(x) = x^x is a continuous function on the real numbers, hence if we calculate the LHL and RHL we would get 1 in both cases. Just try 0.00000001^0.00000001 or -0.0000001^(-0.0000001). LHL = RHL = 1 and also the continuity. I don't get it why this problem poses so many controversy (Don't forget that 0^0 is a completely different thing than the limit as x goes to 0 of x^x)

- 4 years, 11 months ago

Whether definable as a resemblance would make people to consider for whether 0^0 =1. A hand phone calculator gives 0^0 = 1 and I read from a member in Brilliant that many mathematicians take 0^0 = 1. The only thing is people don't know why.

Whenever we have sufficient good reasons, we can validate this.

- 4 years, 11 months ago

Yes you are right, but the problem was asking for the limit not the 0^0 case.

- 4 years, 11 months ago

The limit asked is 1 and not doubted here. Only 0^0 is of actual discussions. Since I disagree with limit x --> 0 for x^x and 0^0 are completely different, I wrote to answer.

- 4 years, 11 months ago

x^x has no left hand limit: try (-0.01)^(-0.01), which is invalid, as fractional powers of negative numbers don't exist

- 4 years, 11 months ago

Fractional power of a negative number exists. It just need not be a real number. Thus, $(-0.5)^{-0.5} = 1/(-0.5)^{0.5} = -i\sqrt{2}$. And your problem $(-0.01)^{-0.01} = -\sqrt{100}\left(\cos(\pi/100) - i\sin(\pi/100)\right)$

- 4 years, 11 months ago

This is a good observation. We actually evaluate this as (0.01)^(-0.01) e^ [(j Pi) (-0.01)]

= 1.0471285480508995334645020315281 (Cos 0.01 Pi - j Sin 0.01 Pi)

= 1.0471285480508995334645020315281 (0.99950656036573155700069083670925 - j 0.031410759078128293839183673817829)

= 1.0466118533231172518572375041105 - j 0.032891102546657091868384347944302

- 4 years, 11 months ago

Since 0^0 is not any value like 0/ 0 but only first possible for between 1 and 0, we ought to find the best resemblance.

With 0^0 expressed as 0^(N - N) = 0^N/ 0^N, where N = N, we can see that 0 isn't the best fit but 1.

Not to let 0^0 be doubted as anything, we define 0^0 = 1. The thing is just that we take symbol 0 for something as thought as exact 0 to stand for 0^0 = 1; taking symbol 0 to represent zero is a good and natural thing to do. This shall not mislead when we think carefully. If we want to think that 0 can never be any other thing when taking any index, then it is better not to take the power to zero and quest or define as undefined. This is why I take 0^0 = 1.

To elaborate these zeroes, we can imagine an extreme tiny thing that is effectively nothing, yet still valid but can be thought as zero. d x/ d x is just 1 no matter how nothing is d x. The thing is (d x)^2/ d x or d x/ (d x)^2 and etc cannot be the 0^0 we mean. Since the concept of 0^0 = 1 is good or better than anything or undefined, it should be taken.

Undefined should be applied onto something else that cannot be expressed or assigned with any fixed quantity. In my development of a 3 Dimensional Number Calculator, I think memorizing the total number of such zeroes is very good. 0/ 0 becomes 0^M/ 0^N for 0^(M - N). As we give the same right to 0 as other numbers, we shall find that 0/ 0 and 0^0 are just as usual as any other cases for other numbers! Everything is generalized.

- 4 years, 11 months ago

Python says it converges to 1.

 1 2 3 4 x = 1 while x >= 0.0000000000000001: print x**x x *= 0.1 

Maybe it's not healthy to place faith in limited accuracy floating point numbers but I'm going to trust it anyway.

- 4 years, 11 months ago

In my opinion, the value of $0^0$ is indeterminate.

Reason : If $0^0 = x$, where $x$ is a definite complex number, then $0^1=x \times 0 = 0$. This relation is true for any $x$. Hence, it is not wise to assign a value to $0^0$.

However, just like $\lim_{x \to 0}\frac{\sin(x)}{x} = 1$ though the function itself indeterminate at $x=0$, $\lim_{x \to 0} x^x = 1$.

The limit exists but the function is not defined at $x=0$

- 4 years, 11 months ago

We can't prove in a way that Sin 30 d = 0.5 while also Sin 150 d = 0.5 to imply that 30 d and 150 d are the same thing. By mean of limit, d y/ d x differs from delta y/ delta x because d y/ d x emphasizes a ratio in limit which resemble a quantified value as thought as both d y and d x arrived to zero. Although both 0's of 0^0 are as thought as arrived to zero, there is still a ratio like d y/ d x which can be valid and definable.

By thinking that Sin x = x - x^3/ 3 ! to be divided by x in limit, the ratio is just x/ x = 1 and when x and x continue to divide, they divide in exactly the same proportion to maintain as a ratio of 1. Never be exactly zero but the limit is as thought as zero because we can't describe a limit by not as thought as exact!

- 4 years, 11 months ago

0/ 0 must be indeterminate, otherwise, there is no d y/ d x. Typical example is tangent to a curve or a circle.

0^0 is not 0/ 0. It is something that tends to a limit like d y/ d x which gives a slope of 1, which we could write 0^0.

As thought as exact zeroes but not actually zeroes, a limit is however to mean for exact arrival to zero. We can think of a tiny limit of nothing left to our sight, yet still valid or exists in a sense that we don't see. A slope or ratio is left for people to realize a fixed direction.

- 4 years, 11 months ago

Relatedly, this problem is one of my favorites:

What is $\lim_{ x \rightarrow 0^+ } x ^ { \frac{1} { \log x } }$?

Staff - 4 years, 11 months ago

Okay, I see why this should be interesting, it suggests that $0^0=e$, but of course that's true only as a limiting value for that particular expression. Yet another reason why $0^0$ cant be anything except indeterminate.

- 4 years, 11 months ago

Right, this shows the path dependence. As mentioned above if the path that you take is polynomial then the limiting value is 1. (I believe that if $f', g'$ is finite, the limiting value is 1). However, if you take a different path, then you could get a different answer.

Staff - 4 years, 11 months ago

Even without the limit, does it suggest that $x^{\frac{1}{\log(x)}}=e$, for all $x$ where the function is defined??

- 4 years, 11 months ago

[a^(Log x)]^(1/ Log x) or [x^(Log x)]^(1/ Log x) is to mean for a or x by Calvin, not only e.

This becomes a form of 0^(0-) to mean for any outcome of a or x, following the base of Log mentioned.

- 4 years, 11 months ago

x^(1/ Log x) = x^(Log a) = a = Antilog (Log a) = a, where a can be x; base taken is x for Antilog and Log.

When x <> a, Limit x--> 0 has no effect to a at all. Limit x--> 0 x^(Log a) = a isn't a consequence of a limit.

- 4 years, 11 months ago