The number $ e$

One of the most important limits in the Calculus is

$\displaystyle \lim_{x \to 0} (1 + x)^{\frac{1}{x}} = 2.71828\cdots = e
$

To prove rigorously that such a limit $ e$ exists, is beyond the scope of this book. For the present we shall content ourselves by plotting the locus of the equation

$\displaystyle y = (1 + x)^{\frac{1}{x}}
$

and show graphically that, as $ x \dot= 0$, the function $ (1 + x)^\frac{1}{x}(= y)$ takes on values in the near neighborhood of $ 2.718\dots$, and therefore $ e = 2.718\dots$, approximately.

Figure 3.14: The function $ (1+x)^{1/x}$.
\includegraphics[height=4cm,width=8cm]{limite.eps}

$ x$ -.1 -.001 .001 .01 .1 1 5 10
$ y=(1+x)^{1/x}$ 2.8680 2.7195 2.7169 2.7048 2.5937 2.0000 1.4310 1.0096

As $ x \to 0-$ from the left, $ y$ decreases and approaches $ e$ as a limit. As $ x \to 0+$ from the right, $ y$ increases and also approaches $ e$ as a limit.

As $ x \to \infty$, $ y$ approaches the limit $ 1$; and as $ x \to -1+$ from the right, $ y$ increases without limit.

Natural logarithms are those which have the number $ e$ for base. These logarithms play a very important rĂ´le in mathematics. When the base is not indicated explicitly, the base $ e$ is always understood in what follows in this book. Thus $ \log_e\ v$ is written simply $ \log\ v$ or $ \ln\ v$.

Natural logarithms possess the following characteristic property: If $ x \to 0$ in any way whatever,

$\displaystyle \lim \frac{\log(1 + x)}{x} = \lim \log(1 + x)^{\frac{1}{x}} = \log\ e
= \ln e %
= 1.
$

david joyner 2008-11-22