In Ch.IV, § 73, we proved that \(\{1 + (1/n)\}^{n}\) tends, as \(n \to \infty\), to a limit which we denoted provisionally by \(e\). We shall now identify this limit with the number \(e\) of the preceding sections. We can however establish a more general result, viz. that expressed by the equations \[\begin{equation*} \lim_{n\to\infty} \left(1 + \frac{x}{n}\right)^{n} = \lim_{n\to\infty} \left(1 – \frac{x}{n}\right)^{-n} = e^{x}. \tag{1} \end{equation*}\] As the result is of very great importance, we shall indicate alternative lines of proof.

(1) Since \[\frac{d}{dt} \log(1 + xt) = \frac{x}{1 + xt},\] it follows that \[\lim_{h\to 0} \frac{\log(1 + xh)}{h} = x.\] If we put \(h = 1/\xi\), we see that \[\lim \xi \log\left(1 + \frac{x}{\xi}\right) = x\] as \(\xi \to \infty\) or \(\xi \to -\infty\). Since the exponential function is continuous it follows that \[\left(1 + \frac{x}{\xi}\right)^{\xi} = e^{\xi\log\{1+(x/\xi)\}} \to e^{x}\] as \(\xi \to \infty\) or \(\xi \to -\infty\): *i.e.* that \[\begin{equation*} \lim_{\xi\to\infty} \left(1 + \frac{x}{\xi}\right)^{\xi} = \lim_{\xi\to -\infty} \left(1 + \frac{x}{\xi}\right)^{\xi} = e^{x}. \tag{2} \end{equation*}\]

If we suppose that \(\xi \to \infty\) or \(\xi \to -\infty\) through integral values only, we obtain the result expressed by the equations (1).

(2) If \(n\) is any positive integer, however large, and \(x > 1\), we have \[\int_{1}^{x} \frac{dt}{t^{1+(1/n)}} < \int_{1}^{x} \frac{dt}{t} < \int_{1}^{x} \frac{dt}{t^{1-(1/n)}},\] or \[\begin{equation*} n(1 – x^{-1/n}) < \log x < n(x^{1/n} – 1). \tag{3} \end{equation*}\] Writing \(y\) for \(\log x\), so that \(y\) is positive and \(x = e^{y}\), we obtain, after some simple transformations, \[\begin{equation*} \left(1 + \frac{y}{n}\right)^{n} < x < \left(1 – \frac{y}{n}\right)^{-n}. \tag{4} \end{equation*}\] Now let \[1 + \frac{y}{n} = \eta_{1},\quad 1 – \frac{y}{n} = \frac{1}{\eta_{2}}.\] Then \(0 < \eta_{1} < \eta_{2}\), at any rate for sufficiently large values of \(n\); and, by (9) of § 74, \[\eta_{2}^{n} – \eta_{1}^{n} < n\eta_{2}^{n-1} (\eta_{2} – \eta_{1}) = y^{2}\eta_{2}^{n}/n,\] which evidently tends to \(0\) as \(n \to \infty\). The result now follows from the inequalities (4). The more general result (2) may be proved in the same way, if we replace \(1/n\) by a continuous variable \(h\).

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