Back to e

Above, we defined $e$ as the particular real number satisfying

$$\lim_{\delta\to 0} \frac{e^\delta-1}{\delta} \isdef 1$$

which gave us $(a^x)^\prime = a^x$ when $a=e$. From this expression, we have, as $\delta\to 0$,

$$\begin{eqnarray*} e^\delta - 1 & \rightarrow & \delta \\ \Rightarrow \qquad e^... ... \\ \Rightarrow \qquad e & \rightarrow & (1+\delta)^{1/\delta}, \end{eqnarray*}$$

or

$$\zbox {e \isdef \lim_{\delta\to0} (1+\delta)^{1/\delta}.}$$

This is one way to define $e$. Another way to arrive at the same definition is to ask what logarithmic base $e$ gives that the derivative of $\log_e(x)$ is $1/x$. We denote $\log_e(x)$ by $\ln(x)$.

Numerically, $e$ is a transcendental number (a type of irrational number^3.5), so its decimal expansion never repeats. The initial decimal expansion of $e$ is given by^3.6

$$e = 2.7182818284590452353602874713526624977572470937\ldots\,.$$

Any number of digits can be computed from the formula $(1+\delta)^{1/\delta}$ by making $\delta$ sufficiently small.