If in the first inequality (3) of § 74 we put $$r = 1/(n – 1)$$, $$s = 1/n$$, we see that $(n – 1)(\sqrt[n-1]{\alpha} – 1) > n(\sqrt[n]{\alpha} – 1)$ when $$\alpha > 1$$. Thus if $$\phi(n) = n(\sqrt[n]{\alpha} – 1)$$ then $$\phi(n)$$ decreases steadily as $$n$$ increases. Also $$\phi(n)$$ is always positive. Hence $$\phi(n)$$ tends to a limit $$l$$ as $$n \to \infty$$, and $$l \geq 0$$.

Again if, in the first inequality (7) of § 74, we put $$s = 1/n$$, we obtain $n(\sqrt[n]{\alpha} – 1) > \sqrt[n]{\alpha}\left(1 – \frac{1}{\alpha}\right) > 1 – \frac{1}{\alpha}.$ Thus $$l \geq 1 – (1/\alpha) > 0$$. Hence, if $$\alpha > 1$$, we have $\lim_{n \to \infty} n(\sqrt[n]{\alpha} – 1) = f(\alpha),$ where $$f(\alpha) > 0$$.

Next suppose $$\beta < 1$$, and let $$\beta = 1/\alpha$$; then $$n(\sqrt[n]{\beta} – 1) = -n(\sqrt{\alpha} – 1)/\sqrt[n]{\alpha}$$. Now $$n(\sqrt[n]{\alpha} – 1) \to f(\alpha)$$, and (Exs. XXVII. 10) $\sqrt[n]{\alpha} \to 1.$ Hence, if $$\beta = 1/\alpha < 1$$, we have $n(\sqrt[n]{\beta} – 1) \to -f(\alpha).$ Finally, if $$x = 1$$, then $$n(\sqrt[n]{x} – 1) = 0$$ for all values of $$n$$.

Thus we arrive at the result: the limit $\lim n(\sqrt[n]{x} – 1)$ defines a function of $$x$$ for all positive values of $$x$$. This function $$f(x)$$ possesses the properties $f(1/x) = -f(x),\quad f(1) = 0,$ and is positive or negative according as $$x > 1$$ or $$x < 1$$. Later on we shall be able to identify this function with the Napierian logarithm of $$x$$.

Example. Prove that $$f(xy) = f(x) + f(y)$$. [Use the equations $f(xy) = \lim n({\sqrt[n]{xy}} – 1) = \lim \{n(\sqrt[n]{x} – 1)\sqrt[n]{y} + n(\sqrt[n]{y} – 1)\}.]$