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In What is $\lim_{x \rightarrow 0} x^0$?

There are discussion that the number is 1.

Of course $ 0^{0} $ isn't well defined. $ 0^{x} $ is $0$ and $ x^{0} $ is always 1 for all $x\neq 0$

I'd say the limit will either be 1 or $0$ depending on how we approach it. It seems that if we approach it from $ x^{x} $

However, I want simple proofs.

One proof says that

$ \lim_\limits{x \rightarrow 0} x^{x} = \lim_\limits{x \rightarrow 0} e^{\log( x^{x})} $

$= \lim_\limits{x \rightarrow 0} e^{\log( x^{x})}$

$= e^{\lim_\limits{x \rightarrow 0} \log( x^{x})}$

$= e^{\lim_\limits{x \rightarrow 0} x \cdot \log(x)}$

I am stuck here. That's because $\log(0)$ is not well defined either. So how to show that it's 1?

sai-kartik
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user4951
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  • as $x≠0$, $x^0=1$. You probably asked for $\lim \limits_{x \to 0} x^x =1$ – Alapan Das May 20 '20 at 15:47
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    Usually, $0^0$ is defined as $1$ as that turns out to be a useful definition in most cases. It is $\lim\limits_{(x,y)\to(0,0)}x^y$ that does not exist. – robjohn May 20 '20 at 16:04
  • @robjohn .......Out of curiosity, could you name such instances where this definition is used – Saket Gurjar May 20 '20 at 16:40
  • To name two, power series $\sum\limits_{k=0}^\infty a_kx^k$ and binomial expansions $(1+x)^n=\sum\limits_{k=0}^n\binom{n}{k}x^k$. In both of these, $0^0=1$ is assumed. To name another, when counting functions from $A$ to $B$, we get $|B|^{|A|}$ and when $A=B=\emptyset$, we have $0^0=1$, the empty function. – robjohn May 20 '20 at 16:45
  • $x^x$ is completely irrelevant if you want to evaluate $\lim_{x\to 0}x^0$. $x^0$ is not the same function as $x^x$, so why should they have the same limit? $x^0=1$ for all $x\ne 0$, so the limit is unambiguously $1$. – TonyK May 20 '20 at 17:08
  • @TonyK: I think the confusion is that in a lot of places, it is said that $0^0$ is not well-defined because $\lim\limits_{(x,y)\to(0,0)}x^y$ does not exist. I think a lot of weakly-founded questions (I'm not saying this one is or is not one of them), are posed because of this confusion. Some may think that $\lim\limits_{x\to0}x^x$ is the same thing. – robjohn May 20 '20 at 17:31
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    @robjohn: yes, of course. The intent of my comment was to impress on the OP that the limit under discussion is not a question of opinion; it is well defined and simple to evaluate. – TonyK May 20 '20 at 17:37
  • Perhaps I should move the rant to my answer. I have posted before regarding $0^0$ and $\lim\limits_{(x,y)\to(0,0)}x^y$. It will remain a topic here, even if it does appear in the FAQ. – robjohn May 20 '20 at 17:58
  • Does this answer your question? Zero to the zero power – is $0^0=1$? – Namburu Karthik May 21 '20 at 15:47

3 Answers3

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The argument you are trying to use can be finished by setting $$ \lim_{x \to 0} x^x = \exp\left( \lim_{x \to 0} x \ln(x) \right) = \exp\left( \lim_{x \to 0} \frac{\ln x}{1/x} \right) \overset{*}{=} \exp\left( \lim_{x \to 0} \frac{1/x}{1/x^2} \right) = e^0 = 1, $$ where step (*) follows by L'Hospital's rule.

However, if you are seeking to define $0^0$, note that this may be achieved in a variety of forms. You are choosing $$ 0^0 := \lim_{x \to 0} x^x = 1, $$ but other forms could be chosen, with $$ 0^0 := \lim_{x \to 0} f(x)^{g(x)}, $$ as long as both $f,g \to 0$ as $x \to 0$, potentially giving a different result.

gt6989b
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For $x$ near $0$ (but not in $0$) it is $x^0=1$ (as @Alapan Das commented).

So $\lim\limits_{x\to0}x^0=\lim\limits_{x\to0}1=1$

1123581321
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First of all, $\lim\limits_{x\to0}x^0=\lim\limits_{x\to0}1=1$. On the other hand, we can use the series for $\log(1-x)$ to evaluate $\lim\limits_{x\to0^+}x^x$: $$ \begin{align} \lim_{x\to0^+}x\log(x) &=\lim_{x\to1^-}\,(1-x)\log(1-x)\tag1\\[6pt] &=\lim_{x\to1^-}(x-1)\sum_{k=1}^\infty\frac{x^k}k\tag2\\ &=\lim_{x\to1^-}\left(-x+\sum_{k=2}^\infty x^k\left(\frac1{k-1}-\frac1k\right)\right)\tag3\\ &=-1+\sum_{k=2}^\infty\left(\frac1{k-1}-\frac1k\right)\tag4\\[3pt] &=-1+1\tag5\\[12pt] &=0\tag6 \end{align} $$ Explanation:
$(1)$: substitute $x\mapsto1-x$
$(2)$: use the series for $\log(1-x)$
$(3)$: multiply the series by $x-1$
$(4)$: take the limit
$(5)$: telescoping sum
$(6)$: simplify

Therefore, $$ \lim_{x\to0^+}x^x=1\tag7 $$

robjohn
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