75. The limit of $n(\sqrt[n]{x}-1)$

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 $\varphi (n)=n(\sqrt[n]{\alpha }–1)$ then $\varphi (n)$ decreases steadily as $n$ increases. Also $\varphi (n)$ is always positive. Hence $\varphi (n)$ tends to a limit $l$ as $n\to \mathrm{\infty }$, and $l\ge 0$.

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 $l\ge 1–(1/\alpha )>0$. Hence, if $\alpha >1$, we have $$\underset{n\to \mathrm{\infty }}{lim}n(\sqrt[n]{\alpha }–1)=f(\alpha ),$$ 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 $$limn(\sqrt[n]{x}–1)$$ defines a function of $x$ for all positive values of $x$. This function $f(x)$ possesses the properties $$f(1/x)=-f(x),f(1)=0,$$ 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 $$f(xy)=limn(\sqrt[n]{xy}–1)=lim\{n(\sqrt[n]{x}–1)\sqrt[n]{y}+n(\sqrt[n]{y}–1)\}.]$$