How come $\lim_{x\to0}$ $x^x = 1$?

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 $x = 0$?

Solution 1:

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.

Solution 2:

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".