1
$\begingroup$

I know that $0^0$ is one of the seven common indeterminate forms of limits, and I found on wikipedia two very simple examples in which one limit equates to 1, and the other to 0. I also saw here: Prove that $0^0 = 1$ using binomial theorem that you can define $0^0$ as 1 if you'd like.

Even so, I was curious, so I did some work and seemingly demonstrated that $0^0$ always equals 1.

My Work:

$$y=\lim_{x\rightarrow0^+}{(x^x)}$$

$$\ln{y} = \lim_{x\rightarrow0^+}{(x\ln{x})} $$

$$\ln{y}= \lim_{x\rightarrow0^+}{\frac{\ln{x}}{x^{-1}}} = -\frac{∞}{∞} $$

$\implies$ Use L'Hôpital's Rule

$$\ln{y}=\lim_{x\rightarrow0^+}\frac{x^{-1}}{-x^{-2}} $$ $$\ln{y}=\lim_{x\rightarrow0^+} -x = 0$$ $$y = e^{0} = 1$$

What is wrong with this work? Does it have something to do with using $x^x$ rather than $f(x)^{g(x)}$? Or does it have something to do with using operations inside limits? If not, why is $0^0$ considered indeterminate at all?

$\endgroup$
11
  • 6
    $\begingroup$ Yes, using $x^x$ rather than $f(x)^{g(x)}$ is exactly what you're doing wrong. Imagine doing it with $f(x) = x, g(x) =0$ or the other way around, and see what you get. $\endgroup$
    – Arthur
    Commented Dec 30, 2015 at 16:15
  • 2
    $\begingroup$ Indeed. Consider $\lim_{x\to0+}x^0$ and $\lim_{x\to0+}0^x$. You're really concerned with the question of whether $$\lim_{(x,y)\to(0+,0+)}x^y$$ exists. $\endgroup$
    – MPW
    Commented Dec 30, 2015 at 16:18
  • 2
    $\begingroup$ When you move the logarithm inside the limit you have already assumed that the limit of $x^x$ exists. $\endgroup$
    – Kayle of the Creeks
    Commented Dec 30, 2015 at 16:20
  • 1
    $\begingroup$ I deleted my post because apparently people disagreed, but I have often seen $0^0$ to be defined as $1$, particularly in combinatorics. This is a matter of convention though, and the limit will always be undefined for the reason above by MPW. $\endgroup$
    – pancini
    Commented Dec 30, 2015 at 16:20
  • 1
    $\begingroup$ @Elliot: Yes, in combinatorics $0^0$ is defined to be $1$. In analysis, not so much. $\endgroup$
    – user856
    Commented Dec 30, 2015 at 16:26

6 Answers 6

Reset to default
8
$\begingroup$

Someone said that $0^0=1$ is correct, and got a flood of downvotes and a comment saying it was simply wrong. I think that someone, me for example, should point out that while saying $0^0=1$ is correct is an exaggeration, calling that "simply wrong" isn't quite right either. There are many contexts in which $0^0=1$ is the standard convention.

Two examples. First, power series. If we say $f(t)=\sum_{n=0}^\infty a_nt^n$ that's supposed to entail that $f(0)=a_0$. But $f(0)=a_0$ depends on the convention that $0^0=1$.

Second, elementary set theory: Say $|A|$ is the cardinality of $A$. The cardinality of the set off all functions from $A$ to $B$ should be $|B|^{|A|}$. Now what if $A=B=\emptyset$? There as well we want to say $0^0=1$; otherwise we could just say the cardinality of the set of all maps was $|B|^{|A|}$ unless $A$ and $B$ are both empty.

(Yes, there is exactly one function $f:\emptyset\to\emptyset$...)

Edit: Seems to be a popular answer, but I just realized that it really doesn't address what the OP said. For the record, of course the OP is nonetheless wrong in claiming to have proved that $0^0=1$. It's often left undefined, and in any case one does not prove definitions...

$\endgroup$
2
  • $\begingroup$ Indeed, it's called an Empty Function. $\endgroup$
    – user236182
    Commented Dec 30, 2015 at 16:40
  • $\begingroup$ To add to the answer, OP only proves that $\lim_{x\to 0^+} x^x$ is consistent with $0^0=1$; but it doesn't mean something else can't be inconsistent with $0^0=1$ (and indeed $\lim_{x\to 0^+}0^x=0\neq 1$). $\endgroup$
    – user236182
    Commented Dec 30, 2015 at 16:53
1
$\begingroup$

The limit of the function $y^x$ at $[0,0]$ does not exist. In fact it depends on the path used to approach $0$.

$0^0$ is defined to be $1$.

$\endgroup$
2
1
$\begingroup$

As user72694 and addy2012 point out, here is an example for which $f(x) \neq g(x)$.

Consider

$$ \lim_{t \to \infty} \left( e^{-t}\right)^\frac{1}{t} \to 0^0$$

But

$$\lim_{t \to \infty} \left( e^{-t}\right)^\frac{1}{t} = \lim_{t \to \infty} \left( e^{-t \frac{1}{t}}\right) = \lim_{t \to \infty} \left( e^{-1} \right) = \frac{1}{e} \neq 1$$

$\endgroup$
0
$\begingroup$

As I think you correctly guessed, one considers the quantities to be tending to zero independently when one works with an indeterminate form, including the one of type $0^0$. If they are the same you indeed get 1, otherwise the answer may vary.

$\endgroup$
0
$\begingroup$

$0^0=1$ is a convention. You can nevertheless justify it by $$0^0=|\mathcal F(\emptyset, \emptyset)|=1$$ since $|\mathcal F(A,B)|=|B|^{|A|}$ if $|A|,|B|<\infty $.

But you have other "proof" that justify this convention. For exemple, for extending the fact that $a^0=1$ for all $a\in \mathbb R$ (i.e. prolonge $x\longmapsto x^0$ to $1$ in $x=0$ by continuity).

However, you can also make the convention that $0^0=0$ (depending on the context), but it's not often used.

I recall that $\mathcal F(A,B)=\{f:A\longrightarrow B\mid f\text{ is a function}\}$.

$\endgroup$
1
  • $\begingroup$ There is no need to prove that $0^0 = 1$ because this follows directly from the definition $x^n = x \cdots x$ ($n$ $x$'s) and the empty product rule. $\endgroup$
    – Mark
    Commented Feb 12, 2017 at 23:17
-1
$\begingroup$

In here, I wrote a detailed explanation on why $0^0=1$ is convenient to use, and why $0^0$ is undefined.

Copying the text, we have

  1. Write a polynomial, for example $x^3-5x^2+7x+2$. If we write this polynomial as $x^3-5x^2+7x^1+2x^0$, it will give consistency and make it easier to deal with certain notations.

  2. We have that $\lim_{x\to 0+}x^x = 1$ This can be proven easily by taking the natural log and applying L'Hopital's Rule.

  3. From $\frac{d}{dx}\left(x^n\right) = nx^{n-1}$, set $n=1$. This gives that $1 = x^{0}$. Now plug in $x=0$, giving $0^0=1$.

  4. How many functions $f : X \rightarrow Y$ are there if $X = Y = \phi$? There is only one such function. Also the number of functions is given by $0^0$ as well, which means that $0^0=1$.

Now for reasons that $0^0$ is still undefined, copying the text again,

  1. Cauchy, who is a pioneer of analysis, wrote in 1821 that $0^0$ is undefined.

  2. Libri, in 1830, wrote a paper that "proved" $0^0=1$, which was very flawed. Few years later, Mobius wrote a paper in support of Libri, and claimed the following.

If $\lim_{x\to 0^+} f(x) = \lim_{x\to 0^+} g(x) = 0$, we have $\lim_{x\to 0^+}f(x)^{g(x)} = 1$.

This was disproved by a mathematician, with its counterexample being $f(x)=a^{-\frac{1}{x}}, g(x)=x$ for $a>1$.

Again, as noted here, it is convention to leave $0^0$ as undefined.

Of course, in combinatorics, it's very useful to leave (and define) $0^0=1$.

$\endgroup$
2
  • $\begingroup$ Indeed any power series gives the same example. I didn't notice that when I wrote the post on AoPS four months ago :( $\endgroup$
    – rkm0959
    Commented Dec 30, 2015 at 16:42
  • $\begingroup$ Cauchy did not write that $0^0$ is undefined, the history is more subtle than that. He did include it in a list of "indeterminate forms" but it is hard to draw an unambiguous conclusion from that due to the lack of a formal definition. Limit arguments are irrelevant unless one accepts the "continuity rule" that you may only define $f(p)$ if $f$ is continuous at $p$. This unstated but implicitly used rule, combined with Cauchy's results, led to the conclusion that $0^0$ should be left undefined. However, by now this rule has been abandoned over a century ago. $\endgroup$
    – Mark
    Commented Feb 12, 2017 at 23:12

Not the answer you're looking for? Browse other questions tagged .