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What is the value of the exponential function $y=e^{x^{x}-1}$ at $x=0$? I graphed the function on desmos and the value at $x=0$ is $1$. However, I do not know how to show this analytically.

OR.
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okabe42
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    Strictly speaking, that function is normally left undefined at $x=0$. The thing is that $x^x$ is normally defined as $e^{x\ln(x)}$. and The $\ln$ is normally undefined at $x=0$. On the other hand one can compute the limit of that function when $x\to0$. – OR. Jun 24 '17 at 18:50
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    To compute the limit you can write it as before $e^{e^{x\ln(x)}-1}$. We first compute the limit of $x\ln(x)=\frac{\ln(x)}{1/x}$ as $x\to0$. You can use L'Hospital to get that it is the same as the limit of $\frac{1/x}{-1/x^2}=-x$. Which is $0$. Therefore $e^{x\ln(x)}\to1$. This gives that $e^{e^{x\ln(x)}-1}\to1$. – OR. Jun 24 '17 at 18:54
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    @MariePierredeLeTetou If you are going to fully explain the question, please just leave an answer. – Franklin Pezzuti Dyer Jun 24 '17 at 19:22
  • @IshKaul If you found any of these answers helpful, you should accept one of them. – Franklin Pezzuti Dyer Jun 27 '17 at 17:35

4 Answers4

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If you try to evaluate it at $x=0$, you end up getting $$e^{0^0-1}$$ Whose value is not immediately apparent. However, by convention, $0^0$ is often said to be equal to $1$. This is because $$\lim_{x\to 0} x^x=1$$ And if you observe the following graph of $y=x^x$, you can see that this is so:

enter image description here

The value you wish to know does not, strictly speaking, exist, but if you use the limit to let $0^0=1$, then the value is $$e^{0^0-1}$$ $$e^{1-1}$$ $$e^{0}$$ $$1$$ And so the limit of your function at that point is equal to $1$.

Franklin Pezzuti Dyer
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    I think the definition $0^0=1$ is often used because of discrete mathematics (there is one function from the empty set to the empty set) and so you can write a polynomial as $\sum_{i=0}^n a_i x^i$, not because of the limit of $x^x$. – Paul Jun 24 '17 at 22:39
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    The argument you cite for $0^0=1$ is precisely the argument that says it shouldn't be defined: specifically, $$\lim_{\substack{x \to 0^+ \ y \to 0}} x^y$$ is undefined. Restricting attention specifically to the case of $x=y$ as a suggestion for defining $0^0$ is the same mistake that lead people to think about assigning $0/0 = 1$. But for this problem specifically, which is explicitly $x^x$, the convention of continuous extension makes sense. The usual reasons for considering $0^0=1$, incidentally, have nothing to do with this situation. –  Jun 25 '17 at 03:24
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This function is not defined at $0$, but you can compute the limit $\lim_{x\to0}e^{x^x-1}=e^{\lim_{x\to0}x^x-1}$. It happens that$$\lim_{x\to0}x^x=\lim_{x\to0}e^{x\log x}=e^{\lim_{x\to0}x\log x}$$and that\begin{align}\lim_{x\to0}x\log x&=\lim_{x\to0}\frac{\log x}{\frac1x}\\&=\lim_{x\to0}\frac{\frac1x}{-\frac1{x^2}}\\&=-\lim_{x\to0}x\\&=0.\end{align}So, $\lim_{x\to0}x^x=e^0=1$ and therefore $\lim_{x\to0}e^{x^x-1}=e^0=1$.

José Carlos Santos
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Sometimes we express $0^0 = 1$, then we have $e^{0^0-1}=e^{1-1}=e^0=1$. However, normally $x^x$ is undefined at $x=0$. So we take the limit, $ \lim_{x\to 0} (e^{x^x-1}) $. Substitute $x \mapsto x^x$ (see below),

$$ \lim_{x\to 1} (e^{x-1}) = e^0 = 1 $$

Here I show that $x^x \to 1$ as $x \to 0$

\begin{align} y &= \lim_{x \to 0} x^x \\ \ln y &= \lim_{x \to 0} x\ln x \\ &= \lim_{x\to 0} \frac{\ln x}{\frac 1 x} \\ &= \lim_{x \to 0} \frac{ \frac 1 x}{\frac{-1}{x^2}} \\ \ln y & = \lim_{x \to 0} (-x) = 0 \\ y &= 1 \end{align}

Dando18
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We know that $\displaystyle e^{x}$ is a differentiable and continuous function.

So the limit is really dependent on $x^{x}$ value as it approaches $0$.

We can rewrite $x^{x}$ as $e^{x\ln x}$. We need to compute $\displaystyle \lim_{x \to 0}e^{x \ln x}=e^{\displaystyle \lim_{x \to 0}x \ln x}=e^{\displaystyle \lim_{x \to 0}\frac{\ln x}{\frac1x}}=e^{\displaystyle \lim_{x \to 0}\frac{\frac{1}{x}}{-\frac{1}{x^2}}}=e^{\displaystyle \lim_{x \to 0} -x}=e^0=1$.

Therefore, we know that $x^{x}-1$ approaches $0$.

And thus $e^{x^x-1}$ approaches $1$.

Saketh Malyala
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