Definition of $Nth$ root:
$3rd$ order inverse group $1$ hyperoperation.
Division is how many times you can subtract a certain divisor from the dividend before it becomes negative.
Likewise Nth root is the result of repeated division by a certain divisor before it becomes $1$ or a decimal. The number of times you divide it before it becomes a decimal is the index.
Ex: $\sqrt [3]{8} = (8/2)/2$
Is the zeroth root even defined and if so what is $\sqrt [0]{x}$
On
The $n^\text{th}$ root of a real number $x$ is $$x^{1/n}$$
If $n=0$ then $1/0$ is undefined, so there is no such thing as the $0^{\text{th}}$ root.
On
The $n$-th root is the inverse of the operation $x \mapsto x^n$. (Let's take $x>0$ to avoid complications.)
For $n=0$, the operation $x \mapsto x^0$ is not invertible, since it is a constant function.
So, there is no $0$-th root operation.
On
No this is usually not defined. One definition of the $n$th root is $$ \sqrt[n]{x} = x^{\frac{1}{n}}. $$ So for a fixed $x>1$ you see that $$ \lim_{n\to 0^+} x^{\frac{1}{n}} = \infty. $$
On
We can look for the value of zeroth root of 2 using limits.
Using trusty WolframAlpha to do this for us we arrive at the conclusion that the limit of 2^(1\n) as n approaches 0 is either 0 or infinity depending on the direction from which we arrive to 0 which means that the zeroth root is undefined.
On
If we reform this, it could be represented as $X^{\frac{1}{0}}$, so when used in standard functional mathematics, it would be undefined, because $\frac{1}{0}$ is undefined.
However, if used in the calculation of a limit and the function reduces to this form after substitution, I'd think this would be considered an indeterminate form, because not enough information is given to determine the original limit. It could be any number. It would end up then looking like $1^\infty$, which would itself be an indeterminate form.
Hint: $$\sqrt[n]{x}=x^{1/n},$$ if $n=0$ then ...