## 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 $$\sqrt{x}$$, $$\sqrt{x}$$, $$\sqrt{x}$$, … 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.