# 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 the term $$x^x$$ must be $$x>0$$ in the other case we get a complex number. – Dr. Sonnhard Graubner Dec 24 '18 at 15:32
• 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 $\mathbb{R}$? – Aniruddha Deshmukh Dec 24 '18 at 15:32
• 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. – SmileyCraft Dec 24 '18 at 15:33
• $x^y$ is not defined if$x<0$ and$y$ is not an integer. – Bernard Dec 24 '18 at 15:40
• Are you trying to typeset limits from above/below? Just want to be sure before I edit with corrected LaTeX. – The Pointer Dec 24 '18 at 15:57

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!

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.

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