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I was always taught by my professors when I was doing Calculus that we only have limit (two-sided) of a function if the function itself is approachable from the left and the right. (Note that, here I'm talking about the limit in Real set).

Geometrically, I still can't process why

$$\lim_{x \to 0^-} x^x = 1,$$

in spite of the fact that algebraically we can show that the limit above is approaching $1$ from the left by rewriting the limit as

$$\lim_{x \to 0^-} e^{\ln{\left(x^x\right)}}$$

and reducing it with L'hôpital's rule we get:

$$\lim_{x \to 0^-} e^{-x} = 1.$$

However, I graphed the function on Geogebra and we also know that there are some value for $x<0$ in $\Bbb R$ (not all values) that make $y$ isn't available in the real set. So, why is the function said to be approachable from the left (geometrically)? We know if $x<0$ and when $y$ isn't available in $\Bbb R$, $y\in \Bbb C$, right? Is it because the continuity of the left part of the value of the limit of the function doesn't matter or what?

P.S. Here, I also provide what Wolfram Alpha says about the limit:

enter image description here

user516076
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    $\lim_{x \to 0^+} x^x =1$ but how do you even define $x^x$ for negative $x$? Especially negative irrational $x$? – Henry Nov 03 '21 at 01:36
  • @Henry I'm sorry I'm lost. I mean, we still have the possibility $y$ is available in $\Bbb R$ for example $x=-1$ – user516076 Nov 03 '21 at 01:38
  • "... I still can't process why..." Why do you think $\lim_{x \to 0^-} x^x = 1$? This identity is not even false. Where is this from? –  Nov 03 '21 at 01:39
  • What do you think $x^x$ is when $x=-\frac12$? When $x=-\frac13$? When $x=-\frac1\pi$? You need it defined for all $x$ near $0$ to be able to take a limit – Henry Nov 03 '21 at 01:41
  • @LouisPan As I said before that we only have limit (two-sided)... Most people believe that the value of the two-sided limit is one (in my Maths group). Actually, I also want to confirm this. – user516076 Nov 03 '21 at 01:42
  • Sorry, I don't understand. You said "I still can't process why $\lim_{x \to 0^-} x^x = 1$"; and I asked where you saw this. This is a "wrong" identity. Or you thought you should have $\lim_{x \to 0} x^x = 1$ and hence the question? –  Nov 03 '21 at 01:45
  • @Henry Still lost sorry. Are you saying the limit isn't 1. Or the left-sided limit doesn't exist? – user516076 Nov 03 '21 at 01:45
  • limit is just approaching the value from all directions possible, if you can only approach it from one direction then that's the direction that matters for the limit. Also you know that $x^x$ for $x < 0$ requires mapping to $\mathbb{C}$ yet you still tried to plot it on Geogebra ...? – okzoomer Nov 03 '21 at 01:46
  • @okzoomer Sorry I didn't explain it in detail. But I'm talking about the value in Real set. – user516076 Nov 03 '21 at 01:49
  • I am saying the limit naturally does not exist using real numbers as $x^x$ is undefined for most negative non-integers, but a suitable treatment might. One approach (which could cause other problems) might be to look at $|x|^{-|x|} e^{-i \pi x}$ when $x$ is negative and see its limit goes to $1$ as $x \to 0^-$. This requires complex numbers. – Henry Nov 03 '21 at 01:49
  • then for $f:\mathbb{R}\to\mathbb{R}$ defined over $(a, b)$ you only look at one-sided limit if you approach $a$ or $b$ – okzoomer Nov 03 '21 at 01:51
  • Related: https://math.stackexchange.com/questions/637280/limit-of-sqrt-x-as-x-approaches-0 – Hans Lundmark Nov 03 '21 at 06:15

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In the world of real analysis, the function $f(x)=x^x$ is defined only for positive real numbers. Its domain is $(0,\infty)$. So it does not make sense to write $\lim_{x \to 0^-} x^x$ and $\lim_{x \to 0}f(x)$ does not exist$\dagger$. On the other hand, $$ \lim_{x \to 0^+} x^x = \lim_{x \to 0^+} e^{x\ln x}=1 $$ where we use the definition $x^x:=e^{x\ln(x)}$ for $x>0$.

In complex analysis, one could extend the function for negative values of $x$ (or even complex values in general). What one has then is $$ x^x=e^{x\text{Ln}(x)},\quad x<0 $$ where $\text{Ln}$ takes a branch that contains the negative x-axis.

$\dagger$ Sometimes, people write $\displaystyle\lim_{x\to a;x\in S}f(x)$ where $S\subset\mathbb{R}$ is the domain of $f$ and $a$ is an accumulation point of $S$. When the context is clear, one may write $\lim_{x\to a}f(x)$, omitting $x\in S$. In the above example, if one writes $\lim_{x\to 0}x^x=1$, one really means $\lim_{x\to 0^+}x^x=1$.