## Calculus 8th Edition

One of the functions may not be defined at x=6, but the product might be. This gives us an idea to set up a counterexample: $f(x)=(x-6),$ $g(x)=\displaystyle \frac{1}{(x-6)}$ $\displaystyle \lim_{x\rightarrow 6}[f(x)g(x)]=\lim_{x\rightarrow 6}\frac{x-6}{x-6}=\lim_{x\rightarrow 6}1=1$ which does not equal $f(6)g(6)$ (g(6) is not defined). The problem statement is false.