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? 

Let $x<0$. It holds that $x^x=e^{x\log x} = e^{x (\log(-x)+\pi i)} = e^{x\log(-x)}(\cos(\pi x) + i \sin (\pi x))$. And I think you can complete the details.
