$\lim\limits_{x \to 0^+}\frac{1}{x}=+\infty$
$\lim\limits_{x \to 0^-}\frac{1}{x}=-\infty$
$\lim\limits_{x \to 0^+}x^x=$?
$\lim\limits_{x \to 0^-}x^x=$?
UPDATE:
According to this graph:
$\lim\limits_{x \to 0^+}x^x=1$
$\lim\limits_{x \to 0^-}x^x=1$
Does that make sense?
Note that $x^x=e^{x\log x}$ and the exponential function is continuous, so it suffices to know what $x\log x$ does as $x \to 0^{\pm}$.
To work out $\displaystyle \lim_{x \to 0^+} x^x$ you can apply l'Hôpital's rule to the following: $$\lim_{x \to 0^+} x\log x = \lim_{x \to 0^+} \frac{\log x}{1/x} = \underbrace{\cdots \cdots \cdots}_{\text{you do this bit}}$$
As for $\displaystyle \lim_{x \to 0^-} x^x$, notice that when $x<0$ $$\log(x) = \log |x| + i\arg(x) = \log (-x)+i\pi$$ Now you get something slightly different: $$\lim_{x \to 0^-} x\log x = \lim_{x \to 0^-} \frac{\log (-x) + i\pi}{1/x} = \underbrace{\cdots \cdots \cdots}_{\text{you do this bit}}$$
Aside: for the $x \to 0^-$ case you have to pick a branch of the logarithm. I chose $\pi$, but strictly speaking any $(2k+1)\pi$ for $k \in \mathbb{Z}$ would work (and give you the same answer).
I would try a logarithmic approach. What is $$L:=\lim_{x \to 0}\ln(x^x)=\lim_{x \to 0} x\ln(x)?$$ Take that value and evaluate $e^L$.
$$\lim_{x\rightarrow{0}}x^x=\exp{\lim_{x\rightarrow{0}}x\ln{x}}$$
Write $x\ln{x}$ as $\frac{\ln{x}}{1/x}$ and apply L'Hopital's rule.
