Investigate \(\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}\) using a table of numbers.
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.
\(\underrightarrow{x~\hbox{approaches 0 from the left}}\) | \(\underleftarrow{x~\hbox{approaches 0 from the right}}\) | ||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|
\(x\) (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 |
Table 3 suggests that \(\lim\limits_{x\rightarrow 0^{-}}f(x)=1\) and \(\lim\limits_{x\rightarrow 0^{+}}f(x)=1\). This suggests that \(\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}=1\).