## Calculus: Early Transcendentals 8th Edition

The statement is false. Here's an example to disprove: We have $f(x)=x-6$ and $g(x)=\frac{1}{x-6}$ So, $$\lim\limits_{x\to6}[f(x)g(x)]$$$$=\lim\limits_{x\to6}\Big[(x-6)\times(\frac{1}{x-6})\Big]$$$$=\lim\limits_{x\to6}1$$$$=1$$ Obviously, here $\lim\limits_{x\to6}[f(x)g(x)]$ exists and equals 1. However, when we consider $\frac{1}{x-6}$, we find that the defining range for it is $x\in R$ except $x=6$. in other words, $\frac{1}{x-6}$ is not defined when $x=6$. That means here $g(6)$ is not defined. So, $f(6)g(6)$ is not defined and not calculable. Which means $\lim\limits_{x\to6}[f(x)g(x)]\ne f(6)g(6)$