How to derive $\lim_\limits{x \rightarrow 0} x^0$? In What is $\lim_{x \rightarrow 0} x^0$?
There are discussion that the number is 1.
Of course $ 0^{0} $ isn't well defined.  $ 0^{x} $ is $0$ and $ x^{0} $ is always 1 for all $x\neq 0$
I'd say the limit will either be 1 or $0$ depending on how we approach it. It seems that if we approach it from $ x^{x} $
However, I want simple proofs.
One proof says that
$ \lim_\limits{x \rightarrow 0}  x^{x} = \lim_\limits{x \rightarrow 0}   e^{\log( x^{x})}   $
$= \lim_\limits{x \rightarrow 0} e^{\log( x^{x})}$
$= e^{\lim_\limits{x \rightarrow 0} \log( x^{x})}$
$= e^{\lim_\limits{x \rightarrow 0} x \cdot \log(x)}$
I am stuck here. That's because $\log(0)$ is not well defined either. So how to show that it's 1?
 A: The argument you are trying to use can be finished by setting
$$
\lim_{x \to 0} x^x
 = \exp\left( \lim_{x \to 0} x \ln(x) \right)
 = \exp\left( \lim_{x \to 0} \frac{\ln x}{1/x} \right)
 \overset{*}{=}
   \exp\left( \lim_{x \to 0} \frac{1/x}{1/x^2} \right)
 = e^0 = 1,
$$
where step (*) follows by L'Hospital's rule.

However, if you are seeking to define $0^0$, note that this may be achieved in a variety of forms. You are choosing
$$
0^0 := \lim_{x \to 0} x^x = 1,
$$
but other forms could be chosen, with
$$
0^0 := \lim_{x \to 0} f(x)^{g(x)},
$$
as long as both $f,g \to 0$ as $x \to 0$, potentially giving a different result.
A: For $x$ near $0$ (but not in $0$) it is $x^0=1$ (as @Alapan Das commented). 
So $\lim\limits_{x\to0}x^0=\lim\limits_{x\to0}1=1$
A: First of all, $\lim\limits_{x\to0}x^0=\lim\limits_{x\to0}1=1$. On the other hand, we can use the series for $\log(1-x)$ to evaluate $\lim\limits_{x\to0^+}x^x$:
$$
\begin{align}
\lim_{x\to0^+}x\log(x)
&=\lim_{x\to1^-}\,(1-x)\log(1-x)\tag1\\[6pt]
&=\lim_{x\to1^-}(x-1)\sum_{k=1}^\infty\frac{x^k}k\tag2\\
&=\lim_{x\to1^-}\left(-x+\sum_{k=2}^\infty x^k\left(\frac1{k-1}-\frac1k\right)\right)\tag3\\
&=-1+\sum_{k=2}^\infty\left(\frac1{k-1}-\frac1k\right)\tag4\\[3pt]
&=-1+1\tag5\\[12pt]
&=0\tag6
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto1-x$
$(2)$: use the series for $\log(1-x)$
$(3)$: multiply the series by $x-1$
$(4)$: take the limit
$(5)$: telescoping sum
$(6)$: simplify
Therefore,
$$
\lim_{x\to0^+}x^x=1\tag7
$$
