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 $$h$$.