Does the value of the $\lim_{x \to 0-} x^x = 1$? I have the following attempt.
Let $x=-y$ then ${y \to 0+}$ as ${x \to 0-}$.
So, $\displaystyle\lim_{x \to 0-} {x}^{x}$=  $\displaystyle\lim_{y \to 0+} {(-y)}^{(-y)} = \displaystyle\lim_{y \to 0+} \dfrac{1}{{(-y)}^{y}}= \displaystyle\lim_{y \to 0+} \dfrac{1}{{(-1)}^{y}.{y}^{y}}=\displaystyle\lim_{y \to 0+} \dfrac{1}{{y}^{y}}$
Now as, $\displaystyle\lim_{y \to 0+} y^y =\displaystyle\lim_{y \to 0+} {e}^{y\ln{y}}
= {e}^{\displaystyle\lim_{y \to 0+} y\ln{y}}={e}^{\displaystyle\lim_{y \to 0+} \frac{\ln{y}}{\frac{1}{y}}} = {e}^{\displaystyle\lim_{y \to 0+} \frac{\frac{1}{y}}{{-\frac{1}{y^2}}}} 
= {e}^{\displaystyle\lim_{y \to 0+} {-y}}=e^{0}=1$ 
Hence $\displaystyle\lim_{y \to 0+} \dfrac{1}{{y}^{y}}=\dfrac{1}{1}=1$
So, $\displaystyle\lim_{x \to 0-} {x}^{x}=1$
Is it correct? 

For complex values of $z$ and $w$, we have by definition
$$\begin{align}
z^w&=e^{w\log(z)}\\\\
&=e^{w\text{Log}(|z|)+iw\arg(z)}\tag1
\end{align}$$
where $\text{Log}$ is the logarithm function of real variables and $\arg(z)$ is the multi-valued argument of $z$.
Using $(1)$ reveals for $x\in \mathbb{R}$ and $x<0$
$$\begin{align}
\lim_{x\to 0^-}x^x&=\lim_{x\to 0^-}e^{x\text{Log}(|x|)+ix\arg(x)}\\\\
&=\lim_{x\to 0^-}x^{|x|}e^{ix(2n+1)\pi}\\\\
&=1
\end{align}$$
as was to be shown!
