199–201. The behaviour of $\log x$ as $x$ tends to infinity or to zero

199. The manner in which $\log x$ tends to infinity with $x$.

It will be remembered that in Ex. XXXVI. 6 we defined certain different ways in which a function of $x$ may tend to infinity with $x$, distinguishing between functions which, when $x$ is large, are of the first, second, third, … orders of greatness. A function $f(x)$ was said to be of the $k$th order of greatness when $f(x)/x^k$ tends to a limit different from zero as $x$ tends to infinity.

It is easy to define a whole series of functions which tend to infinity with $x$, but whose order of greatness is smaller than the first. Thus $\sqrt{x}$, $\sqrt[3]{x}$, $\sqrt[4]{x}$, … are such functions. We may say generally that $x^{\alpha }$, where $\alpha$ is any positive rational number, is of the $\alpha$th order of greatness when $x$ is large. We may suppose $\alpha$ as small as we please, e.g. less than $.0000001$. And it might be thought that by giving $\alpha$ all possible values we should exhaust the possible ‘orders of infinity’ of $f(x)$. At any rate it might be supposed that if $f(x)$ tends to infinity with $x$, however slowly, we could always find a value of $\alpha$ so small that $x^{\alpha }$ would tend to infinity more slowly still; and, conversely, that if $f(x)$ tends to infinity with $x$, however rapidly, we could always find a value of $\alpha$ so great that $x^{\alpha }$ would tend to infinity more rapidly still.

Perhaps the most interesting feature of the function $\log x$ is its behaviour as $x$ tends to infinity. It shows that the presupposition stated above, which seems so natural, is unfounded. The logarithm of $x$ tends to infinity with $x$, but more slowly than any positive power of $x$, integral or fractional. In other words $\log x\to \mathrm{\infty }$ but $$\frac{\log x}{x^{\alpha }}\to 0$$ for all positive values of $\alpha$. This fact is sometimes expressed loosely by saying that the ‘order of infinity of $\log x$ is infinitely small’; but the reader will hardly require at this stage to be warned against such modes of expression.

200. Proof that $(\log x)/x^{\alpha }\to 0$ as $x\to \mathrm{\infty }$.

Let $\beta$ be any positive number. Then $1/t<1/t^{1-\beta }$ when $t>1$, and so $$\log x={\int }_1^x\frac{dt}{t}<{\int }_1^x\frac{dt}{t^{1-\beta }},$$ or $$\log x<(x^{\beta }–1)/\beta <x^{\beta }/\beta ,$$ when $x>1$. Now if $\alpha$ is any positive number we can choose a smaller positive value of $\beta$. And then $$0<(\log x)/x^{\alpha }<x^{\beta -\alpha }/\beta (x>1).$$ But, since $\alpha >\beta$, $x^{\beta -\alpha }/\beta \to 0$ as $x\to \mathrm{\infty }$, and therefore $$(\log x)/x^{\alpha }\to 0.$$

201. The behaviour of $\log x$ as $x\to +0$.

Since $$(\log x)/x^{\alpha }=-y^{\alpha }\log y$$ if $x=1/y$, it follows from the theorem proved above that $$\underset{y\to +0}{lim}{y}^{\alpha }\log y=-\underset{x\to +\mathrm{\infty }}{lim}(\log x)/x^{\alpha }=0.$$ Thus $\log x$ tends to $-\mathrm{\infty }$ and $\log (1/x)=-\log x$ to $\mathrm{\infty }$ as $x$ tends to zero by positive values, but $\log (1/x)$ tends to $\mathrm{\infty }$ more slowly than any positive power of $1/x$, integral or fractional.