Prove that $ \lim_{x\to 0}x^x=1$. Prove that $\displaystyle \lim_{x\to 0}x^x=1$ for $\delta-\epsilon.$
We have to $|x|<\delta_{\epsilon}$ what will be chosen later.
$$\begin{align*}|x^x-1|<\epsilon &\Leftrightarrow -\epsilon<x^x-1<\epsilon\\ &\Leftrightarrow -\epsilon+1<x^x<\epsilon+1\\ &\Leftrightarrow \ln(-\epsilon+1)<x\ln x<\ln(\epsilon+1)\end{align*}$$
Hence I do not know how to continue, i.e. how to choose my $ \delta $. Someone help me, thanks.
 A: The solution involves the Lambert function. See https://en.wikipedia.org/wiki/Lambert_W_function 
First of all, $x^x$ is defined for $x>0$ and has a minimum at $x=1/e$ 
Secondly  the right inequality of the expression $1-\epsilon < x^x <1+\epsilon$ is true as long as if $\delta \leq 1$, and so it is true also for $\delta \leq 1/e < 1$
We focus on the left inequality and call $k=1-\epsilon \leq 1$ for simplicity.
We are looking for a value $\delta \leq 1/e$  such as $\delta^\delta >k$, and since $x^x$ decreases when $x \leq 1/e$ it suffices look for a $\delta$ that $\delta ^\delta = k$
The Lambert function is defined as $W(x)=f^{-1}(x)$ where $f(x)= x \cdot e^x$ and its domain is $[-1/e, \infty)$
Now we have some properties:
(1) $W(f(x))=f^{-1}(f(x))=x \hskip{1cm}$So $W(f(x))=W(xe^x)=x$ 
(2) $f(W(x))=f (f^{-1}(x))=x \hskip{1cm}$ So $x =f(W(x))=W(x)e^{W(x)} \Rightarrow e^{W(x)}=\frac{x}{W(x)}$ 
Now we focus in our search: $\delta^\delta = k \Leftrightarrow \delta \ln \delta = \ln k \Leftrightarrow e^{\ln \delta}\ln \delta =\ln k$ 
And applying Lambert: $W(e^{\ln \delta}\ln \delta) = W(\ln(k))$ 
If $z=\ln \delta$, the previous equation translates into $W(ze^z)=W(\ln(k))$ 
And since $W(ze^z)=z$ (prop 1), we have $\ln \delta = z=W(\ln(k)) \Rightarrow \delta = e^{W(\ln(k))}$. 
But $e^{W(b)}=\frac{b}{W(b)}$ (prop 2)$ \Rightarrow \delta =\frac{\ln k}{W(\ln k)}$
So the solution is:
$\delta = \min \{1/e, \frac{\ln k}{W(\ln k)}\}$ where $k=1-\epsilon$
A: As suggested, I scrapped my original answer in favor of an $\delta,\epsilon$ approach, where L'Hopital's rule is not used.
I am re-interpreting the problem to be : 
To prove: $\lim_{x \to 0^+} x^x = 1.$ 
This is because, for any $\delta > 0$, there exists an $x$ such that $-\delta < x < 0$ and $x^x$ is undefined in Real Analysis.

For $~x > 0,~$ let $~f(x) = x\log x~$ and let $~g(x) = x^x.$ 
Then $~g(x) = e^{[f(x)]}.$
$\underline{\text{Lemma 1 :}}$
For $~0 < x < \dfrac{1}{e}~, f(x)~$ is a strictly decreasing function. 
Proof: 
$f'(x) = 1 + \log(x) < 1 + (-1) < 0.$ 
Thus, $f(x)$ is strictly decreasing throughout the interval.
$\underline{\text{Lemma 2 :}}$
Let $~\langle a_n\rangle~$ be defined by
$a_n = \dfrac{3^n}{n} ~: ~n \in \Bbb{Z^+}.$ 
Then $~\langle a_n\rangle~$ is a strictly increasing sequence that grows unbounded. 
Proof:
$\displaystyle \frac{a_{n+1}}{a_n} = 
\frac{3}{(n+1)/n} = \frac{3n}{n+1} > 1 ~: ~\forall n \in \Bbb{Z^+}.$ 
Therefore, $~\langle a_n\rangle~$ is a strictly increasing sequence. 
Further, for $\displaystyle n \geq 3, 
\frac{3}{(n+1)/n} > \frac{3}{3/2} = 2.$ 
Therefore, for $\displaystyle n \geq 3, 
\frac{a_{n+1}}{a_n} > 2$. 
Therefore, for $\displaystyle n \geq 4,
a_n > 2^{(n-3)}a_3 = 2^{(n-3)} \times 9.$ 
Clearly, as $n \to \infty, 2^{(n-3)}$ grows unbounded.
$\underline{\text{Lemma 3 :}}$
$x < y \implies e^x < e^y.$ 
Proof: 
$h(x) = e^x \implies h'(x) = e^x > 0.$ 
Thus, $h(x)$ is a strictly increasing function.

The challenge is that given any $\epsilon > 0,$ 
$\delta > 0$ must be found so that 
$0 < x < \delta \implies 0 \leq |g(x) - 1| < \epsilon.$ 
This will be true $\iff 1 - \epsilon < g(x) < 1 + \epsilon.$
In constructing a $\delta, \epsilon$ proof, $\epsilon$ will be arbitrarily restricted to be less than $1$, so that $0 < (1 - \epsilon) < 1.$
This is justified, since if a suitable $\delta$ can be found for a specific $\epsilon_1 < 1$, such a $\delta$ would also be suitable for any $\epsilon > \epsilon_1$.
Given $\epsilon$ such that $0 < \epsilon < 1$, use the following procedure to construct $\delta$:
By Lemma 2, $n \in \Bbb{Z^+}$ may be chosen so that
$$\frac{n}{3^n} |\log(1/3)| < 
|\log (1-\epsilon)|.\tag 1$$
Choose $\displaystyle \delta = \frac{1}{3^n} \implies \delta \times |\log(\delta)| = 
\frac{1}{3^n} \times n |\log (1/3)|.$
Note that for any $x$ such that $0 < x \leq \delta$, 
$\log (x) < 0 \implies x \log x < 0 \implies g(x) = x^x < 1 < (1 + \epsilon).$
Further, by (1) above, $\delta |\log (\delta)| < |\log(1-\epsilon)| \implies \delta \log (\delta) > \log(1 - \epsilon)$.
Also, since $\displaystyle \delta = \frac{1}{3^n} \leq \frac{1}{3} < \frac{1}{e}$, 
Lemma 1 guarantees that for all $x$ such that $0 < x < \delta, ~~$ you will have that $~~x \log (x) > \log (1-\epsilon).$
Invoking Lemma 3, this implies that $ g(x) = x^x > 1 - \epsilon.$
