#### Answer

False.

#### Work Step by Step

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.