Answer
$ - \frac{1}{3}$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 3} \frac{{\ln \left( {x/3} \right)}}{{3 - x}} \cr
& {\text{Use the quotient property of limits}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to 3} \left[ {\ln \left( {x/3} \right)} \right]}}{{\mathop {\lim }\limits_{x \to 3} \left( {3 - x} \right)}} \cr
& {\text{Evaluate the limit when }}x \to 3 \cr
& = \frac{{\ln \left( {3/3} \right)}}{{3 - 3}} = \frac{0}{0} \cr
& {\text{The limit is an indeterminate form type }}\frac{0}{0},{\text{ so we can apply}} \cr
& {\text{the L'Hospital's Rule}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 3} \frac{{\ln \left( {x/3} \right)}}{{3 - x}} = \mathop {\lim }\limits_{x \to 3} \frac{{\frac{d}{{dx}}\left[ {\ln \left( {x/3} \right)} \right]}}{{\frac{d}{{dx}}\left[ {3 - x} \right]}} \cr
& = \mathop {\lim }\limits_{x \to 3} \frac{{\frac{{1/3}}{{x/3}}}}{{ - 1}} \cr
& = - \mathop {\lim }\limits_{x \to 3} \frac{1}{x} \cr
& {\text{Evaluate the limit when }}x \to 3 \cr
& = - \frac{1}{3} \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to 3} \frac{{\ln \left( {x/3} \right)}}{{3 - x}} = - \frac{1}{3} \cr} $$