What is the solution to $\lim_{x\to0^-} x^x?$ $$\lim_{x\to0^-} x^x.$$
Symbolab says it's 1. My answer sheet says it does not exist. Which is correct, and can you explain the solution?
 A: The term $$x^x$$ is only defined for all reals if $$x>0$$ holds.
A: EDIT:  I see the question has changed slightly since I began typing the first draft of this.  I added clarification to answer the actual question, and left the rest of the post in its entirety in case it helps a future viewer.

$\lim_{x \to 0} x^x$ does not exist.
$\lim_{x \to 0^+} x^x = 1$.
Note the subtle difference.  In the limit that exists, we approach $x$ only from the right.  The limit that doesn't exist is the two-sided limit.  This is because $x^x$ is not defined for $x < 0$, therefore we cannot approach $x=0$ from the left.  Note that this also implies that $\lim_{x \to 0^-} x^x$ does not exist.
As for why the defined limit is 1:
$$x^x = e^{\ln x^x} = e^{x \ln x}$$
Therefore,
$$\lim_{x \to 0^+} x^x = 
  \lim_{x \to 0^+} e^{x \ln x} = e^{\lim_{x \to 0^+} x\ln x}.$$
We can push the limit inside the exponent like that because limits can be pushed into and pulled out of continuous functions.
Anyway, we can use l'Hôpital to evaluate $\lim_{x \to 0^+} x\ln x$.
\begin{align}
  \lim_{x \to 0^+} x \ln x &= \lim_{x \to 0^+} \frac{\ln x}{1/x}\\[0.3cm]
    &= \lim_{x \to 0^+} \frac{1/x}{-1/x^2}\\[0.3cm]
    &= \lim_{x \to 0^+} (-x)\\[0.3cm]
    &= 0
\end{align}
Therefore $\lim_{x \to 0^+} x^x = e^0 = 1$.
A: The problem is that $x^x$ is ill-defined when $x < 0$, which is the domain your limit is considering. For example, $(-0.5)^{-0.5}$ is not defined. $(-1/3)^{-1/3}$ exists, but $(-1/4)^{-1/4}$ does not, and so on. In fact there are negative values of $x$ arbitrarily close to $0$ so that $x^x$ is not defined. By the definition of the limit, in such a case the limit cannot be defined.
A: Is it considered over real domain? If so, then the limit does not exists, because $$x^x=e^{xlog(x)}$$ and one cannot take a logarithm of a non-positive number.
