Is $\sqrt[0]{0}$ defined? Is $\sqrt[0]{0}$ defined?
How about the limit? $\lim_{x\to 0}{x^{\frac{1}{x}}}$
Is $\sqrt[0]{n}\,\,\,\,n\neq0$ any different? 
 A: We have $$\lim_{x\to0^+}x^{1/x}=0$$ since $x^{1/x}$ is not defined for all negative $x$.
This is because as $x\to0$, $\frac1x\to\infty$ and multiplying a number $<1$ by itself $n$ times (as $n\to\infty$) gets you to zero.
Notice that we are writing $x\to0$, not $x=0$, as $x^{1/x}=0^\infty$ is indeterminate.
A: The limit dose not exist because for negative values of $x$, $x^{1/x}$  is not defined.
A: Let $L = \lim_{x \to 0}\root{x}\of {x}$. Then, taking logarithm on both sides,
$$\begin{aligned}
\ln {L} = \lim_{x \to 0}\frac{\ln{x}}{x}
\end{aligned}$$
This quantity is not defined, that is, it is an indeterminate of the form $\frac{\infty}{0}$.
A: No. It is undefined, and here's why.
You write $\sqrt[0]{n}$, but I will write it as $\sqrt[0]{x}$ with respect to the limit (because it has a variable $x$).

Regarding the zeroth root,$$\begin{align}\text{Let }\;\;\,y&=\sqrt[0]{x}, \\ \text{then }\;y^0&=x.\end{align}$$

Since anything raised to the power of $0$ is $1$, it follows then that $x=1$. $$\therefore \sqrt[0]x\text{ is undefined}\Leftrightarrow x\neq 1.\tag{$\therefore \sqrt[0]{0}$ is undefined.}$$ But $y$ could be any number, too, so this means $\sqrt[0]{x}$ has no definite value, even if $x\neq 1$.


$$\therefore \sqrt[0]{x}\text{ is undefined } \forall x.$$




Regarding the limit,
$$\Lambda=\lim_{x\to0}x^{\frac 1x}$$ this is of course equal to $0$. $\overbrace{\text{Since $0^n=0$ for all $n\neq 0$,}}^{\large Statement \ (1)}$ then as $x\to 0$, it follows that $x^{\frac 1x}\to 0$, and by definition of a limit, we get that $\Lambda = 0$.

But this does not mean that $1\div 0$ is defined — it is not; division by zero is undefined, regardless of the value of the numerator. We just have to consider the limit, $$\begin{align}\Lambda_0=\lim_{x\to0}\frac 1x&=\infty, \\ \because \lim_{x\to\infty}\frac 1x&=0.\end{align}$$ Therefore, when evaluating the limit $\Lambda$, we notice that $1\div x\to \infty$.


And thus, by Statement $(1)$, it is fully clear that $\Lambda = 0$; id est, $$\lim_{x\to 0}\frac 1x=0.$$



If you are still unsure and/or confused, take a look at the best answer (in my opinion).
