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Disclaimer: I am completely and utterly new here. Please openly correct me if I'm doing anything wrong.

I received the following question on a recent calc test:
$$\lim_{x \to - \infty} x^x=?$$ $$a) 1 \quad b) \infty \quad c) 0 \quad d) DNE$$

I selected $d) DNE$ as my answer as I believed the function $f(x) = x^x$ has a domain of $x \in (0, \infty)$, and any negative value will cause the function to oscillate between positive and negative values and result in discontinuities. I was sincerely surprised when the test was returned and the correct answer was $c) 0$.

My instructor's argument was that the function did not have to be continuous for the limit to exist, despite it being outside of the domain.

My argument was that for certain negative real numbers of the form $-\frac{2a+1}{2b}$ for $a, b \in \mathbb{N}$, the function will yield imaginary results of the form: $$\frac{1}{\sqrt[2b]{\left(-\frac{2a+1}{2b}\right)^{2a+1}}}$$. My friend also had a counter-example where $\left(-\frac{1}{2}\right)^{-\frac{1}{2}}$ does not exist, but $\left(-\frac{2}{4}\right)^{-\frac{2}{4}}$ does.

I have done some research on my own, and I believe by the epsilon-delta definition of a limit, this limit does not exist as this limit does not satisfy that for all real $x$ in the needed range, $\left|f(x) - L\right| < \epsilon$.

Please lend help!

Edit: is there a (more) rigorous way to show that the limit does not exist?

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    $\begingroup$ For the reason you pointed out, negative numbers can only really be raised to integer powers. If your instructor meant the domain to be integers, then they're correct. $\endgroup$
    – Zoe Allen
    Mar 11, 2023 at 3:24
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    $\begingroup$ @ZoeAllen My instructor did not specify the domain of the function. The given problem, in its entirety, has been presented above. $\endgroup$
    – Zero G
    Mar 11, 2023 at 3:27
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    $\begingroup$ @Seeker well, the part where the prof failed to define it, etc. was definitely spot-on (I upvoted it too) $\endgroup$
    – obscurans
    Mar 11, 2023 at 4:04
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    $\begingroup$ Welcome to MSE. Excellent first question $\endgroup$ Mar 11, 2023 at 8:01
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    $\begingroup$ I also want to chime in that this is an excellent example of a first question. I agree with the assessment of others that, unless you're a math major or a 3rd year+ physics major, you almost certainly don't have the math background to get the limit to be zero. I'm not entirely sure I fully understand, because analysis isn't my strength. If this is a high school class, MSE is collectively face-palming at your teacher. $\endgroup$ Mar 11, 2023 at 10:14

2 Answers 2

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Since you marked this as in calculus, not complex analysis, you're correct. For fractional negative $x$, you're definitely going to get complex results, and at the scope of calculus, those numbers don't exist, so clearly if the function doesn't even exist, what limit can?

If you're fully allowing complex numbers to show up, then you end up with a horrendous mess of intricate details such as branch cuts (choosing the complex argument), multivalued functions (or choose all!), and so on. The limit can (with much conceptual difficulty and definitions you haven't learned) be shown to be $0$ - see @MarkViola's answer.

Your instructor is doing the physics/engineering thing of saying because $$x^y=\exp(y\ln(x))$$ completely ignoring all the mathematical subtleties (in particular, that $\ln(x)$ typically behaves the worst right at negative reals - it's the cut), then the limit can be "eventually" worked out.

When the initial steps already broke spectactularly.


For a rigorous argument that the limit fails to exist in real analysis:

  1. $x^x$ fails to exist for all $\{-n-1/2\}, n\in\mathbb{N}$.
  2. Therefore the function is undefined for arbitrarily large negative $x$.
  3. Therefore the function cannot have any limit at all.
  4. Suppose it did, and look at the definition for a limit at $-\infty$: $$\forall\epsilon>0\;\exists\delta\in\mathbb{R}\quad|\quad x<\delta\Rightarrow\left|f(x)-L\right|<\epsilon$$ but you can always exhibit a value $x<\delta$ where $f(x)$ is undefined, let alone within $\epsilon$ of any purported limit. $\Rightarrow\Leftarrow$
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  • $\begingroup$ In the complex analysis interpretation, is it even a mess or is it just ill defined? there isn't a way to raise a negative number to an irrational power is there? $\endgroup$
    – Zoe Allen
    Mar 11, 2023 at 3:29
  • $\begingroup$ You have to do some form of functional continuation of the exponentiation function to allow for fully complex base and argument, and it's technically possible, with a bunch of cases that won't work. $\endgroup$
    – obscurans
    Mar 11, 2023 at 3:30
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    $\begingroup$ Wiki details some possibilities. The interesting thing? The price is that you break specifically these identities such as $\ln(a^b)=b\ln(a)$ no matter what you do. $\endgroup$
    – obscurans
    Mar 11, 2023 at 3:36
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    $\begingroup$ The argument of $x$, $\arg(x)$, does not impact the magnitude of $x^x$, $\left|x^x\right|=\left|e^{x\log(|x|)+ix\arg(x)}\right|=e^{x\log(|x|)}$ for real values of $x$. Therefore, irrespective of the definition of the complex logarithm, the coveted limit is indeed $0$. I posted a solution accordingly. $\endgroup$
    – Mark Viola
    Mar 11, 2023 at 4:18
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    $\begingroup$ @MarkViola I agree you have the answer correctly. My claim stands that to a calculus student, there remains at least a year of instruction before they can deal with the complex logarithm and its details (multiple definitions!) enough to be able to understand that kind of proof. $\endgroup$
    – obscurans
    Mar 11, 2023 at 4:21
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Unless your class has covered the complex logarithm, which is multi-valued on $\mathbb{C}$, your answer is correct (i.e., in the context of real analysis, the limit does not exist).

However, in the context of complex analysis we have for $x\in \mathbb{R}$

$$\begin{align} \left|x^x\right|&=\left|e^{x\log(|x|)+ix\arg(x)}\right|\\\\ &=e^{x\log(|x|)} \end{align}$$

which does go to zero as $x\to -\infty$. Therefore, irrespective of the choice of the definition of $\arg(x)$, the coveted limit is indeed $0$.

NOTE: Note that here, $\arg(x)$ denotes the multi-valued argument.

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    $\begingroup$ We note that $x^x$ is multi-valued when $x<0$ and $x \not\in \mathbb Z$. However, all the values have the same modulus. That is, $|x^x|$ is the same for all values of $x^x$. And (calculation) $\lim_{x\to-\infty} |x^x| = 0$. $\endgroup$
    – GEdgar
    Mar 11, 2023 at 15:08
  • $\begingroup$ @GEdgar Indeed. I thought that was clear in the solution herein. $\endgroup$
    – Mark Viola
    Mar 11, 2023 at 16:21

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