Example

Investigate $\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}$ using a table of numbers.

Solution The domain of the function $f(x)=\dfrac{\sin x}{x}$ is $\{x~|~x\neq 0\}$ . So, $f$ is defined everywhere in an open interval containing 0, except for 0.

We create Table 3, by investigating one-sided limits of $\dfrac{\sin x}{x}$ as $x$ approaches 0, choosing numbers (in radians) slightly less than 0 and numbers slightly greater than 0.

TABLE 3

$\underrightarrow{x~\hbox{approaches 0 from the left}}$								$\underleftarrow{x~\hbox{approaches 0 from the right}}$
$x$ (in radians)	-0.02	-0.01	0.005	$\rightarrow$	0	$\leftarrow$	0.005	0.01	0.02
$f(x) =\dfrac{\sin x}{x}$	0.99993	0.99998	0.999996	$f(x)$ approaches 1			0.999996	0.99998	0.99993

${f(x)=}\dfrac{\sin {x}}{{x}}$ is an even function, so the bottom row of Table 3 is symmetric about $x=0$ .

Table 3 suggests that $\lim\limits_{x\rightarrow 0^{-}}f(x)=1$ and $\lim\limits_{x\rightarrow 0^{+}}f(x)=1$ . This suggests that $\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}=1$ .