Is $x^x$ continuous on negative integers?

Question

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.

I could be wrong about this, does someone have some input for me?

sure! https://math.stackexchange.com/questions/1551470/domain-of-xx/2846434#2846434 — B flat, Jul 12 '18 at 07:02
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." — B flat, Jul 12 '18 at 07:03
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. — , Jul 12 '18 at 07:25
For a function to be continuous at a point a then the limit must at least be defined there right? So it would seem we should look at the definition of limit at a point c. The definition I have is " Assuming that f(x) is defined for all x in some open interval containing a, except possibly at a, then we say the limit of (x) as x approaches L is...". I'm not sure how you can get around this definition since continuity at c requires the limit to exists at c. The limit definition cannot even be applied without the function being defined an on open interval around c. Am I missing something? — B flat, Jul 12 '18 at 07:30
@MichaelMcCain: What you wrote is true if the domain of the function is an interval. Have you seen the $\varepsilon$-$\delta$ definitions of limits and continuity? — , Jul 12 '18 at 07:35
Yes, actually the definition stated the first part of the epsilon-delta definition of the limit at a point c. I left off the rest. This definition is standard for limits in Calculus I courses. But maybe it's not loose enough for advanced thought? — B flat, Jul 12 '18 at 07:38
I am using the Briggs/Cochran text used across the nation. The Stewart text takes the same approach with the Epsilon-delta definition of the limit. It starts.. "Let f be a function defined on some open interval that contains the number a, except possibly at a itself. Then..." — B flat, Jul 12 '18 at 07:41
@MichaelMcCain: There you go. Your definition starts off with $f$ being defined on some open interval. When the domain is more exotic, then one requires an $\varepsilon$-$\delta$ definition of continuity that does not rely on the $\varepsilon$-$\delta$ definition of limit. See the first answer in: http://math.stackexchange.com/questions/1815425/continuity-at-isolated-point. Note, however, that this is beyond the level of Stewart. I don't know the Briggs/Cochran text and have never seen it used in my nation. — , Jul 12 '18 at 07:58
I see! Now that you mention it in do recall having this conversation in my real analysis course. I would have liked to see how Stewart would have defined continuity in his texts with epsilon delta in a way that avoids the apparent contraction with the first definition of continuity in his text. Thank you! — B flat, Jul 12 '18 at 08:06
@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: $f(x)=\sqrt{x^3-x^2}$. — , Jul 12 '18 at 08:13
Hmmm.... It still seems somewhat inconsistent. So the limit does not exist on the negative integers but the function is continuous on the negative integers? This doesn’t make sense to me. It seems that both definitions should NOT have mention of c being defined on an open interval correct? — B flat, Jul 12 '18 at 08:22
Follow up question... doesn't this mean that the function f(x)={1 for all integers, and undefined non-intergers} is differentiable everywhere it is defined? It seems that if we remove the condition that the function be defined on open intervals around the points in questions, then we get some strange conclusions that don't seem useful. Am I wrong in my example? — B flat, Jul 12 '18 at 08:30
I'll take a look at the link you sent more closely. It seems the answer lies in there. Thank you for your time. — B flat, Jul 12 '18 at 08:34
@MichaelMcCain , note that differentiablity depends on a limit and not not continuity (although continuity is necessary condition) — ℋolo, Jul 12 '18 at 09:30

score 1 · Answer 1 · answered Jul 12 '18 at 07:23

1

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.

answered Jul 12 '18 at 07:23

Szeto

11,159

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? – B flat Jul 12 '18 at 07:33
This is a single variable calculus I course so we aren’t discussing complex analysis. – B flat Jul 12 '18 at 07:34
@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. – Szeto Jul 12 '18 at 07:43
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. : ) – B flat Jul 12 '18 at 08:00
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@MichaelMcCain You are welcome. Please kindly accept my answer if you find it somewhat useful. – Szeto Jul 12 '18 at 08:06

Is $x^x$ continuous on negative integers?

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