$\lim_{x\to -\infty} x^x$ exists? and if so, what's its value?

Question

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?

Answer 1

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.

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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. $\Rightarrow\Leftarrow$

Answer 2

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.