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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?

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  • $\begingroup$ For the term $$x^x$$ must be $$x>0$$ in the other case we get a complex number. $\endgroup$ – Dr. Sonnhard Graubner Dec 24 '18 at 15:32
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    $\begingroup$ How did you write $\left( -1 \right)^y = 1$? Don't you think if we take a sequence $\dfrac{1}{n}$, $\left( -1 \right)^{\frac{1}{n}}$ may not be defined in $\mathbb{R}$? $\endgroup$ – Aniruddha Deshmukh Dec 24 '18 at 15:32
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    $\begingroup$ It depends on your definition of $a^b$ for $a<0$, but using $a^b=e^{b(\ln(-a)+\pi i)}$ your argument should work. $\endgroup$ – SmileyCraft Dec 24 '18 at 15:33
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    $\begingroup$ $x^y$ is not defined if$x<0$ and$y$ is not an integer. $\endgroup$ – Bernard Dec 24 '18 at 15:40
  • $\begingroup$ Are you trying to typeset limits from above/below? Just want to be sure before I edit with corrected LaTeX. $\endgroup$ – The Pointer Dec 24 '18 at 15:57
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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!

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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.

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  • $\begingroup$ Note that for $x<0$, $\log(x)=\text{Log}(|x|)+i(2n+1)\pi$, for $n\in \mathbb{Z}$. $\endgroup$ – Mark Viola Dec 24 '18 at 16:17

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