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 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 $\lim\limits_{x\rightarrow 0}\dfrac{e^{x}-1}{x}=1$.