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:
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 $\lim_{x\to 0^+}x^x=1$.