# Is $x^x$ continuous on negative integers?

In another post someone mentioned that $x^x$ is continuous on the negative integers. All of the definitions for limit that I am aware say that the function has to be defined on an open interval around the point to even talk about a limit there. So I’m not sure how you can say that the function is continuous on the negative integers since there is no open interval around any negative integer where the function is defined.

• Can you pink the other post?
– ℋolo
Jul 12, 2018 at 6:59
• Jul 12, 2018 at 7:02
• he says "f you choose to include all x∈ℤ x ∈ Z with x<0 x < 0 in the domain of your function, your domain/function gets "messed up"; your domain is not an intervall anymore, it is not a differentiable function (it is however still continuous) etc." Jul 12, 2018 at 7:03
• Continuity does not require the function to be defined on an open interval around the point in question. In fact, one can show that if a function $f$ has a domain with an isolated point, then $f$ is continuous at that isolated point. The limit at that point won't exist, though.
– user360359
Jul 12, 2018 at 7:25
• @MichaelMcCain: No problem. One can show that the $\varepsilon$-$\delta$ definition of continuity implies the one in Stewart when the function is on an open interval. By the way, if you'd like an example of a function that looks benign but that has a domain with an isolated point: $f(x)=\sqrt{x^3-x^2}$.
– user360359
Jul 12, 2018 at 8:13

Indeed, the following function is continuous over $\mathbb C$: $$f(z)= \begin{cases} z^z,&z\ne0\\ 1,&z=0\\ \end{cases}$$

Let $z=re^{it}$.

It can be shown that $$f(z)=\text{exp}(e^{it}r\ln r+rt(i\cos t-\sin t))$$

The only ‘problematic point’ is $r=0$.

By showing $$|e^{it}r\ln r|=|r\ln r|\to 0$$ as $r\to 0^+$, one can argue $f(z)$ is continuous everywhere on the complex plane. (Also note that the sum and composition of continuous functions are continuous.)

It does make sense to consider ‘continuity at a point’. Since the function is continuous everywhere, immediately it is continuous over $\mathbb Z^-$ as well.

• Thank you. However, I was only looking at the reals. Is the function continuous at every negative integer over the reals? Maybe I can't around not discussing this over the complex? Jul 12, 2018 at 7:33
• This is a single variable calculus I course so we aren’t discussing complex analysis. Jul 12, 2018 at 7:34
• @MichaelMcCain I looked at all definitions of continuity on Wikipedia and all requires the function to be defined in the neighborhood of $c$ before discussing the continuity at $c$. I do not agree that the function is continuous over the negative integers if you restrict the range of the function to be the reals. Jul 12, 2018 at 7:43
• I see. Thank you! It seems that one must clarify whether we or not we are restricting ourselves to the study of functions of real numbers only. I can tell my Calculus students that later on when they get to complex analysis they will widen our interpretation of functions of real numbers to be functions of complex numbers in which case the same function will be continuous on the negative integers. Thanks for clarifying. It’s been a long time since I’ve looked at my complex analysis textbooks. : ) Jul 12, 2018 at 8:00
• @MichaelMcCain You are welcome. Please kindly accept my answer if you find it somewhat useful. Jul 12, 2018 at 8:06