Continuity of Root Function

Theorem
Let $$n \in \mathbb{N}^*$$ be a non-zero natural number.

Let $$f: \left[{0 \,. \, . \, \infty}\right) \to \mathbb{R}$$ be the real function defined by $$f \left({x}\right) = x^{1/n}$$.

Then $$f$$ is continuous at each $$\xi > 0$$ and continuous on the right at $$\xi = 0$$.

Proof

 * First suppose that $$\xi > 0$$.

Let $$X, Y \in \mathbb{R}$$ such that $$0 < X < \xi < Y$$.

Let $$x \in \mathbb{R}$$ such that $$X < x < Y$$.

From Inequalities Concerning Roots, we have $$X Y^{1/n} \left|{x - \xi}\right| \le n X Y \left|{x^{1/n} - \xi^{1/n}}\right| \le Y X^{1/n} \left|{x - \xi}\right|$$.

Thus $$\frac 1 {n Y} Y^{1/n} \left|{x - \xi}\right| \le \left|{x^{1/n} - \xi^{1/n}}\right| \le \frac 1 {n X} X^{1/n} \left|{x - \xi}\right|$$.

The result follows by applying the Squeeze Theorem.


 * Now we need to show that $$f \left({x}\right) \to 0$$ as $$x \to 0^+$$.

We need to show that $$\forall \epsilon > 0: \exists \delta > 0: x^{1/n} = \left|{x^{1/n} - 0}\right| < \epsilon$$ provided $$0 < x < \delta$$.

Clearly, for any given $$\epsilon$$, we can choose $$\delta = \epsilon^n$$.

Hence the result.