## 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 \(.000\ 000\ 1\). 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 \(\log x \to \infty\) 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 \(x \to \infty\).

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 \[0 < (\log x)/x^{\alpha} < x^{\beta-\alpha}/\beta \quad (x > 1).\] But, since \(\alpha > \beta\), \(x^{\beta-\alpha}/\beta \to 0\) as \(x \to \infty\), 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 \[\lim_{y\to +0} y^{\alpha} \log y = -\lim_{x\to +\infty} (\log x)/x^{\alpha} = 0.\] Thus \(\log x\) tends to \(-\infty\) and \(\log(1/x) = -\log x\) to \(\infty\) as \(x\) tends to zero by positive values, but \(\log(1/x)\) tends to \(\infty\) more slowly than any positive power of \(1/x\), integral or fractional.

$\leftarrow$ 198. The functional equation satisfied by \(\log x\) | Main Page | 202. Scales of infinity $\rightarrow$ |