Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/46

If now $x$ approaches the limit zero,

$${\underset {x=0}{\operatorname {limit} }}\;{\frac {x}{\sin x}}$$

must lie between the constant $1$ and ${\underset {x=0}{\operatorname {limit} }}{\frac {1}{\cos x}}$, which is also $1$.

Therefore ${\underset {x=0}{\operatorname {limit} }}\;{\frac {x}{\sin x}}=1$, or, ${\underset {x=0}{\operatorname {limit} }}\;{\frac {\sin x}{x}}=1$.

Th. III, p. 18

It is interesting to note the behavior of this function from its graph, the locus of equation

$$y={\frac {\sin x}{x}}.$$

Although the function is not defined for $x=0$, yet it is not discontinuous when $x=0$ if we define

23. The number $\mathbf {e}$. One of the most important limits in the Calculus is

$${\underset {x=0}{\operatorname {limit} }}\;(1+x)^{\frac {1}{x}}=2.71828\cdots =e.$$

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

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

and show graphically that, as $x\doteq 0$, the function $(1+x)^{\frac {1}{x}}(=y)$

$x$	$y$	$x$	$y$
10	1.0096
5	1.4310
2	1.7320
1	2.0000
.5	2.2500	—.5	4.0000
.1	2.5937	—.1	2.8680
.01	2.7048	—.01	2.7320
.001	2.7169	—.001	2.7195

takes on values in the near neighborhood of $2.718\cdots$, and therefore $e=2.718\cdots$ approximately.