Continuity of Root Function

Theorem

Let $n \in \N_{>0}$ be a non-zero natural number.

Let $f: \left[{0 .. \infty}\right) \to \R$ 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 $X, Y \in \R$ such that $0 < X < \xi < Y$.

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

From Inequalities Concerning Roots, we have:

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

Thus:

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

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

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

Hence the result.

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 8.15 \ (5)$