The $n$-th root as a continuous inverse: for a natural $n\ge 1$ the map $x\mapsto x^n$ is continuous and strictly increasing on $[0,\mathrm{\infty })$ with image $[0,\mathrm{\infty })$, so its inverse $x\mapsto x^{1\mathrm{/}n}$ is continuous and strictly increasing

Let $n\in \mathbb{N}$ with $n\ge 1$ and put $I:=[0,\mathrm{\infty })$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) and $p:I\to \mathbb{R}$, $p(x):=x^n$ (Integer powers $a^m$). Then:

1. $p$ is continuous on $I$ (Continuity of $f:A\to \mathbb{R}$ at a point of $A$ and on $A$: the $\epsilon$-$\delta$ condition, its agreement with ${\lim }_{x\to c}f(x)=f(c)$ at a limit point, and continuity at an isolated point);
2. $p$ is increasing on $I$ (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $\mathbb{R}$, with the dictionary to monotone sequences), hence injective (Injection, surjection, bijection);
3. $p[I]=I$;
4. consequently the inverse map $g:I\to I$ of $p$ is continuous on $I$ and increasing, and $g(a)=a^{1\mathrm{/}n}$ is the unique nonnegative $n$-th root of $a$ (Existence and uniqueness of $n$-th roots: a unique $a^{1\mathrm{/}n}\ge 0$ with $(a^{1\mathrm{/}n}{)}^n=a$).

So the $n$-th root function is continuous, and it is obtained from Continuous inverse theorem: a continuous injective $f$ on an interval $I$ is a bijection onto the order-convex set $f[I]$, and the inverse $g:f[I]\to I$ is continuous and strictly monotone in the same sense as $f$ rather than from a direct $\epsilon$-$\delta$ estimate.

• Why $n\ge 1$. At $n=0$ the map $x\mapsto x^0$ is constantly $1$ (Integer powers $a^m$), so it is neither injective nor surjective onto $[0,\mathrm{\infty })$, and no inverse exists. Every claim above is stated for $n\ge 1$ and the hypothesis is used in step 1.2.

• Why the domain is $[0,\mathrm{\infty })$ and not $\mathbb{R}$. For even $n$ the map $x\mapsto x^n$ is not injective on $\mathbb{R}$, since $(-x{)}^n=x^n$, so the continuous inverse theorem does not apply there; restricting to the nonnegative reals is what makes it injective, and it is also where Existence and uniqueness of $n$-th roots: a unique $a^{1\mathrm{/}n}\ge 0$ with $(a^{1\mathrm{/}n}{)}^n=a$ provides the roots.

• What is gained over the root theorem alone. Existence and uniqueness of $n$-th roots: a unique $a^{1\mathrm{/}n}\ge 0$ with $(a^{1\mathrm{/}n}{)}^n=a$ produces the number $a^{1\mathrm{/}n}$ for each $a$ separately and says nothing about how it varies with $a$. Claim 4 is the statement that $a\mapsto a^{1\mathrm{/}n}$ is a continuous increasing function, and it comes from the structure of the situation rather than from any estimate on roots.