208. The representation of $e^x$ as a limit

In Ch.IV, § 73, we proved that $\{1+(1/n){\}}^n$ tends, as $n\to \mathrm{\infty }$, 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 $$\begin{array}{}\text{(1)}& \underset{n\to \mathrm{\infty }}{lim}{(1+\frac{x}{n})}^n=\underset{n\to \mathrm{\infty }}{lim}{\left(1–\frac{x}{n}\right)}^{-n}=e^x.\end{array}$$ 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 $$\underset{h\to 0}{lim}\frac{\log (1+xh)}{h}=x.$$ If we put $h=1/\xi$, we see that $$lim\xi \log (1+\frac{x}{\xi })=x$$ as $\xi \to \mathrm{\infty }$ or $\xi \to -\mathrm{\infty }$. Since the exponential function is continuous it follows that $${(1+\frac{x}{\xi })}^{\xi }=e^{\xi \log \{1+(x/\xi )\}}\to e^x$$ as $\xi \to \mathrm{\infty }$ or $\xi \to -\mathrm{\infty }$: i.e. that $$\begin{array}{}\text{(2)}& \underset{\xi \to \mathrm{\infty }}{lim}{(1+\frac{x}{\xi })}^{\xi }=\underset{\xi \to -\mathrm{\infty }}{lim}{(1+\frac{x}{\xi })}^{\xi }=e^x.\end{array}$$

If we suppose that $\xi \to \mathrm{\infty }$ or $\xi \to -\mathrm{\infty }$ 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 $$\begin{array}{}\text{(3)}& n(1–x^{-1/n})<\log x<n(x^{1/n}–1).\end{array}$$ Writing $y$ for $\log x$, so that $y$ is positive and $x=e^y$, we obtain, after some simple transformations, $$\begin{array}{}\text{(4)}& {(1+\frac{y}{n})}^n<x<{\left(1–\frac{y}{n}\right)}^{-n}.\end{array}$$ Now let $$1+\frac{y}{n}={\eta }_1,1–\frac{y}{n}=\frac{1}{{\eta }_2}.$$ 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 $n\to \mathrm{\infty }$. 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 $h$.