Limit of $x^x$ as $x$ tends to $0$ I am trying to solve the following limit:
$$\lim \limits_{x\to0} x^x$$
The only thing that comes to mind is to write $x^x$ as $e^{x\ln{x}}$ and getting the right sided limit would be easy but I don't see how I could get the left sided one seeing that the $\ln$ is not defined for negative numbers.
Is there something I am missing or is there another way to go about it?
P.S.:I don't know anything about derivatives so please keep it to the limits.

First find $\lim\limits_{x\rightarrow 0}x\ln(x)=\lim\limits_{x\rightarrow 0}\frac{\ln(x)}{1/x}$. Using L'Hospital this become $\lim\limits_{x\rightarrow 0}\frac{1/x}{-1/x^{2}}=\lim\limits_{x\rightarrow 0}-x=0$.
So $\lim\limits_{x\rightarrow 0}e^{x\ln(x)}=e^{\lim\limits_{x\rightarrow 0}x\ln(x)}=1$.
(note assuming $x>0$ of course, since $x^{x}$ is not well-defined otherwise)
Also, if you allow $x<0$ but $x$ must be rational only, then the limit do not exist. This can be seen from the fact that $\lim\limits_{x\rightarrow 0}x^{x}=1$ when $x>0$. This means, that there are positive $x$ arbitrarily close to $1$ in any neighbourhood of $0$. Now once you look at negative $x$, then by considering rationals with odd denominator in any neighbourhood of $0$, then the once with even numerator will be positive, and odd numerator will be negative. The positive one will be arbitrary close to $1$, the negative one will be arbitrarily close to $-1$, so there are no limit.
If you allow $x<0$ and $x$ must be rational only, but also allow only a subset of rational such that $x^{x}$ have definite sign, then the limit is either $1$ or $-1$ from the left.
