Value of an exponential function 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.
 A: 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 $\lim_{x\to0}e^{x^x-1}=e^0=1$.
A: 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}
A: 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 $1$.
A: 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:

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 $1$.
