Investigating a Limit Using a Table of Numbers

Investigate \(\lim\limits_{x\rightarrow 0}\dfrac{e^{x}-1}{x}\) using a table of numbers.

Solution The domain of \(f(x)=\dfrac{e^{x}-1}{x}\) is \(\{x| x\neq 0 \}\). So, \(f\) is defined everywhere in an open interval containing the number 0, except for 0.

We create Table 2, investigating the left-hand limit \(\lim\limits_{x\rightarrow 0^{-}}\dfrac{e^{x}-1}{x}\) and the right-hand limit \(\lim\limits_{x\rightarrow 0^{+}}\dfrac{e^{x}-1}{x}\). First, we evaluate \(f\) at numbers less than 0, but close to zero, and then at numbers greater than 0, but close to zero.

TABLE 2
\(\underrightarrow{x~\hbox{approaches 0 from the left}}\) \(\underleftarrow{x~\hbox{approaches 0 from the right}}\)
\(x\) -0.01 -0.001 -0.0001 -0.00001 \(\rightarrow\) 0 \(\leftarrow\) 0.00001 0.0001 0.001 0.01
\(f(x) =\dfrac{e^{x}-1}{x}\) 0.995 0.9995 0.99995 0.999995 \(f(x)\) approaches 1 1.000005 1.00005 1.0005 1.005

Table 2 suggests that \(\lim\limits_{x\rightarrow 0^{-}}\dfrac{e^{x}-1}{x}=1\) and \(\lim\limits_{x\rightarrow 0^{+}}\dfrac{e^{x}-1}{x}=1\). This suggests \(\lim\limits_{x\rightarrow 0}\dfrac{e^{x}-1}{x}=1\).