Where is $x^x$ continuous? The idea of continuity of a function is something I come across quite regularly, but I've never really understood it well. I'm trying to fix that by looking at some interesting functions.

What happens continuity wise for $x&lt;0$ in this function:

$$f(x)=x^x$$

I can't find a quick answer to this. Apologies if this is a duplicate.
I'm looking for a detailed explanation.
As mentioned in another answer, writing $x^x$ is the same as writing $e^{x\log x}$. But complex analysis tells us that there are many possible branches of $\log x$ that you could choose. Namely, you can define it on any simply connected region in $\mathbb C$ not containing zero, and you can add $2\pi i$ as many times as you like. 

Since you're asking about real negative values of $x$, you'd need a branch of the logarithm that avoids this half line in the complex plane. For example, you could choose the branch
$$\log z=\log|z|+i\theta$$
where $-\frac{\pi}{2}&lt;\theta&lt;\frac{3\pi}{2}$. In any such branch, the complex logarithm is analytic and therefore continuous on the negative real half-line.

To conclude, the answer to your question is that $x^x$ is always continuous for $x&lt;0$ provided you've picked a well-defined meaning of the function.

I apologize that this explanation requires that you've seen complex analysis, but this question cannot be adequately answered otherwise.
The quickest and easiest way to make a statement on this function's continuity is to take a derivative. This requires logarithmic differentiation. The derivative is:
$$f'(x) = x^x(\ln(x) + 1)$$

Note that $\ln(x)$ is only valid for positive entries. So, this tells us that $f$ is continuous for positive numbers.