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.

L'Hospital's rule is quickest. I show that other approaches are possible:
For $\ x\in \left[\ \frac{1}{7},\ \frac{1}{6}\ \right),$
$$ \left( \frac{1}{7} \right)^{\frac{1}{6}}<x^x<1.$$
Now using Newton's Binomial expansion,
\begin{align} \left(1-\alpha\right)^\frac{1}{6} = 1 + \frac{1}{6}(-\alpha) + \frac{\left(\frac{1}{6}\right)\left(-\frac{5}{6}\right)}{2!}(-\alpha)^2 + \frac{\left(\frac{1}{6}\right)\left(-\frac{5}{6}\right)\left(-\frac{11}{6}\right)}{3!}(-\alpha)^3+\ldots \\
\\
= 1 - \frac{1}{6}\alpha - \frac{\left(\frac{1}{6}\right)}{2}\frac{\left(\frac{5}{6}\right)}{1} \alpha^2 - \frac{\left(\frac{1}{6}\right)}{3} \frac{\left(\frac{11}{6}\right)}{2} \frac{\left(\frac{5}{6}\right)}{1} \alpha^3 - \ldots\ \\
\\
> 1 - \frac{1}{6}\left(\alpha + \frac{1}{2} \alpha^2 + \frac{1}{3} \alpha^3 + \ldots \right)\\
\\
= 1 - \frac{1}{6}(\ -\ln(1-\alpha)\ ) = 1 + \frac{1}{6}(\ \ln(1-\alpha)\ ).\\
\\
\end{align}
Substituting $\ \alpha = \frac{6}{7},\ $ into the above, we see that
$$ 1+ \frac{1}{6}\ln\left(\frac{1}{7}\right) = 1 - \frac{1}{6}\ln(7) < \left( \frac{1}{7} \right)^{\frac{1}{6}}<x^x<1\quad \forall x\in \left[\ \frac{1}{7},\ \frac{1}{6}\ \right).$$
More generally, for any $\ k \in\mathbb{N},\ $ we have,
$$ 1 - \frac{\ln(k+1)}{k} < \left( \frac{1}{k+1} \right)^{\frac{1}{k}} < x^x < 1 \quad \forall x\in \left[\ \frac{1}{k+1},\ \frac{1}{k}\ \right).$$
Letting $\ k\to\infty\ $ yields the result.
