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Recently I have been researching $x^x$, mainly when $x<0$. All the computing softwares such as Wolfram Alpha don't display the correct graph for negative x values. The actual graph can be seen here: http://peda.com/grafeq/gallery/rogue/xx_exponential.html

I have a found a thread explaining why this happens(Can the graph of $x^x$ have a real-valued plot below zero?). Based on the actual graph, the $\lim_{x\to0^-}$ $x^x$ is equal to $1$ and $-1$. Does that mean that there isn't an actual limit when $x = 0$?

Tim
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  • For negatie non-integers $x$ , $x^x$ is not defined, only the limit from the right makes sense and this limit can be proven to be $1$. – Peter Nov 22 '21 at 14:01

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Note that $f(x)=x^x$ is only defined in real numbers if and only if $x >0$ (or for integer values). So, in a $I^{-}_{(0)}$ you can't calculate the limit. We can evaluate, instead:

$$\lim_{x \to 0^+} x^x = \lim_{x \to 0^+}e^{\ln(x)\cdot x}$$

Letting $t=\frac{1}{x}$, when $x\to 0^+$, $t$ is going to $+\infty$. So, the limit becomes: $$\lim_{t\to +\infty}e^{\frac{\ln(t)}{t}}=e^0=1$$ wehere $\lim_{t\to +\infty}\frac{\ln(t)}{t}$ is due to infinity hierarchy.

Matteo
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  • Not quite true : For negative INTEGER $x$, $x^x$ is defined. But for refuting the limit, it is enough that it is not defined for irrational negative $x$. – Peter Nov 22 '21 at 14:32
  • @Peter: yeah, you are right. For example $(-2)^{-2}$ is defined. I will add this to the answer. Instead, $x^x$ is not defined in irrational, or cases as: $\left(-\frac{1}{2}\right)^{-\frac{1}{2}}$ – Matteo Nov 22 '21 at 14:36
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The text in the picture of what you call the "actual graph" tells you that the graph shown for $x<0$ is that of a function whose domain is only the negative rational numbers with odd denominator when written in lowest terms.

The graph is drawn to suggest an unbroken curve, but it should really be thought of as dense sprinkling of unconnected points.

Nearby points need not be close together: the values at $-998/999$ and $-1000/999$ are far apart: one is near $1$, the other near $-1$. You can find sequences approaching $0$ from the left for which the values approach either $1$ or $-1$, so the limit you seek is indeed undefined.

If you allow $x$ to be a complex number you can make sense of $x^x= e^{x \ln x}$. It will be a "multivalued function".

Ethan Bolker
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For $x<0$, write $x^x=(-1)^x |x|^x$. The key is that $(-1)^x$ is not well-defined. The main idea is that $(-1)^{p/q}$ does make sense if $q$ is an odd integer (for then, $(-1)^{p/q}=(-1)^p$), but for any irrational number $x$, or when $q$ is even (where it is undefined), what happens as a limiting process as $p/q \to x$? What happens as other irrational numbers $y \to x$? There's really no meaningful way to define this.

Now when $x>0$, write $x^x=e^{x\ln x}$. Since $x \mapsto e^x$ is continuous, we can take $\lim_{x\to 0^+} x \ln x$ and then calculate the limit.

ProfOak
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