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 \(\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}=1\).