Why we assume that $a > 0$ for the following law? The law limit for roots:
$$\lim_{x\to a}\sqrt[\leftroot{-2}\uproot{2}n]{x} = \sqrt[\leftroot{-2}\uproot{2}n]{a}$$
(If $n$ is even, we assume that $a > 0$)
However, for the definition of at least a square root, $0$ can also be accepted. Why would we exclude $a = 0$?
 A: You assume that $a>0$ because you're approaching $a$ from two sides. 
$\lim_{x\rightarrow a}\sqrt[n] x=\sqrt[n] a$ means that $$\lim_{x\rightarrow a^-}\sqrt[n] x=\lim_{x\rightarrow a^+}\sqrt[n] x=\sqrt[n] a$$
It is not possible to find $\lim_{x\rightarrow 0^-}\sqrt[n] x$ for even $n$, however $\lim_{x\rightarrow 0^+}\sqrt[n] x=0$.
For another example, try to evaluate $\lim_{x\rightarrow 0^-}2+\frac{|x|}{x}$ and $\lim_{x\rightarrow 0^+}2+\frac{|x|}{x}$.
A: The limit $\lim_ {x\to 0} \sqrt[n] x$ is not defined, since even roots of negative real numbers are not real...  that is $\lim_{x\to 0^-}\sqrt[n] x\not =\lim_{x\to 0^+}\sqrt[n] x$, for $n$ even.
(This phenomenon leads into the study functions of a complex variable, where this issue is dealt with, and you will learn about branch cuts.  There will, in fact, always be points of discontinuity...)
A: The definition of $\lim_{x\to a} f(x) = K$ is: for any $\epsilon > 0$ there is a $\delta$ so that $|x-a|< \delta \implies |f(x)-K| < \epsilon$.
Notice if $f(x) = \sqrt[n]{x}$ where $n$ is even and $a = 0$ this can't actually ever be true!
If $|x-0| < \delta$ then $-\delta < x < \delta$ so it will be possible for $x$ to be negative.  If $x$ is negative it will NOT be the case that $|\sqrt[n]{x} - a| < \epsilon$.
So notice if $a > 0$ and $\epsilon > 0$ and we want to find the $\delta$... Notice that $\delta \le |a|$.  It's not that important and no-one I know ever brings it up but, nonetheless it will be a criterion.
We can get around this by talking only about one-sided limits.
$\lim_{x \to a^+} f(x) = K$ means that for any $\epsilon > 0$ there is a $\delta > 0$ so that if $|x - a| < \delta$ AND $x > a$, or in other words if $0 < x-a < \delta$ then $|f(x) - K| < \epsilon$.
That may be more pertinent.
After all $\lim_{x \to 0^+} \sqrt[2m]{x} = 0$ is true and useful.  But $\lim_{x\to 0^-}\sqrt[2m]{x}$ is completely undefined.
But it is important to realize that $\lim_{x\to a}f(x) = K$ means that BOTH $\lim_{x\to a^+} f(x)$ and $\lim_{x\to a^-} f(x)$, both exist, and both equal $K$.
.... so $\lim_{x\to 0} \sqrt[2m]{x} \ne 0$.  Just isn't.
