Hubbard & Hubbard: domain of $(1+x)^{1/x}$

Question

On page 98 of Hubbard & Hubbard's Vector Calculus, Linear Algebra, and Differential Forms [2nd Ed.], they say of

$$ f(x) = (1+x)^{1/x}, $$

that

although we cannot evaluate the function at 0, the natural domain of the function includes $[-1,0) \cup (0, \infty)$, and 0 is in the closure of that set.

Footnote. We say that the natural domain "includes" $[-1,0) \cup (0, \infty)$ because one might argue that -3 is also in the natural domain; every number, whether positive or negative, has a unique real cube root.

(I don't think the footnote is relevant to my question.)

I'm confused about the inclusion of -1 in the domain. Since

$$ f(-1) = (1 + (-1))^{1/-1} = 0^{-1} = \dfrac{1}{0} , $$

isn't $f$ undefined at $x = -1$? Shouldn't the natural domain be $(-1,0) \cup (0, \infty)$?

Answer 1

From page 92 of the latest edition of the book, Hubbard & Hubbard's Vector Calculus, Linear Algebra, and Differential Forms [5th Ed.]:

For example, does it make sense to speak of $$\lim_{x \to 0} (1+x)^{1/x}?$$ Yes; although we cannot evaluate the function at 0, the natural domain of the function includes $(-1,0) \cup (0,\infty)$, and 0 is in the closure of that set.

I excluded the footnote; it was longer than before, but again, it used $(-1,0) \cup (0,\infty)$ instead of $[-1,0) \cup (0,\infty)$. Therefore, I think we can safely say that the closed bracket in the 2nd edition is an error. The only thing that I still think is curious is that I can't find any mention of it in the 2nd edition errata.