calculus - Limit of $x^x$ without L'Hôpital

I was trying to calculate $\lim_{x \to 0^{+}} x^x$ without L'Hôpital's rule but could not make progress.

My best shot was to show that $\lim_{x \to 0^{+}} x\ln x = 0$ as that would imply the first limit.
Can anyone help me?

Answer

Hint: What else do you know about $\ln x$? How does its growth rate compare to $\frac1x$ or $x$?

Another hint: you can transform this to a $\lim_{y \rightarrow \infty}$ problem i.e. by setting $y=1/x$. I find this much easier to work with. Thinking about things close to $0^+$ is hard.

You will want to use some combination of finding upper / lower bounds, and technically you will apply the squeeze theorem.