TITLE: Why does this limit exist $x^{x}$
QUESTION [5 upvotes]: I was wondering what is the value of $\lim\limits_{x\to 0+} x^{x}$.
Assuming that the limit exists, I could show using the usual logarithm techniques that the limit is $1$. However, I am not able to show that the limit exists. Could some one help on that?
EDIT: It would be so nice if we could do it without l'Hopital...

REPLY [4 votes]: Assume you know 
$$\lim_{n \rightarrow \infty}\sqrt[n]{n}=1$$
Proof:

$$1 \leq \sqrt[n]{n}<\frac{n-2+2\sqrt n}{n}, \,\,\,\,\,\,\,\,\,\, \mathrm{GM \leq AM}$$

Now for each $x$ near zero define $n(x)=[\frac{1}{x}]$ where $[\cdot]$ is the integral part function. Let's call $n(x)$
out of simplicity $n$.
Therefore 
\begin{align}
&\frac{1}{n} \geq x \geq \frac{1}{n+1} \frac{}{} \Rightarrow\\
\left (\frac{1}{n} \right )^\frac{1}{2n}\geq &\left (\frac{1}{n} \right )^{\frac{1}{n+1}} \geq x ^{x}\geq 
\left (\frac{1}{n+1} \right )^\frac{1}{n} \geq  \left (\frac{1}{2n} \right )^\frac{1}{n}&
\end{align}
And because as $x$ tends to zero $n(x)$ tends to infinity taking limits at both sides you are done.