Power Rule for Derivatives/Fractional Index/Proof 1

Proof
Let $n \in \N_{>0}$.

Thus, let $\map f x = x^{1 / n}$.

From the definition of the power to a rational number, or alternatively from the definition of the root of a number, $\map f x$ is defined when $x \ge 0$.

(However, see the special case where $x = 0$.)

From Continuity of Root Function, $\map f x$ is continuous over the open interval $\openint 0 \infty$, but not at $x = 0$ where it is continuous only on the right.

Let $y > x$.

From Inequalities Concerning Roots:
 * $\forall n \in \N_{>0}: X Y^{1 / n} \, \size {x - y} \le n X Y \, \size {x^{1 / n} - y^{1 / n} } \le Y X^{1 / n} \, \size {x - y}$

where $x, y \in \closedint X Y$.

Setting $X = x$ and $Y = y$, this reduces (after algebra) to:


 * $\dfrac 1 {n y} y^{1 / n} \le \dfrac {y^{1 / n} - x^{1 / n} } {y - x} \le \dfrac 1 {n x} x^{1 / n}$

From the Squeeze Theorem for Functions, it follows that:
 * $\ds \lim_{y \mathop \to x^+} \dfrac {y^{1 / n} - x^{1 / n} } {y - x} = \dfrac 1 {n x} x^{1 / n} = \dfrac 1 n x^{\dfrac 1 n - 1}$

A similar argument shows that the left hand limit is the same.

Thus the result holds for $\map f x = x^{1 / n}$.