Answer
$$\frac{1}{e}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to e} \frac{{\ln x - 1}}{{x - e}} \cr
& {\text{rewrite}} \cr
& \mathop {\lim }\limits_{x \to e} \frac{{\ln x - 1}}{{x - e}} = \mathop {\lim }\limits_{x \to e} \frac{{\ln x - \ln \left( e \right)}}{{x - e}} \cr
& {\text{using the definition of the derivative }}f'\left( a \right) = \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}} \cr
& {\text{comparing }}\mathop {\lim }\limits_{x \to e} \frac{{\ln x - \ln \left( e \right)}}{{x - e}}{\text{ with the definition of the derivative}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\lim }\limits_{x \to e} \frac{{\ln x - \ln \left( e \right)}}{{x - e}}:f'\left( a \right) = \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\lim }\limits_{x \to e} \frac{{\ln x - \ln \left( e \right)}}{{x - e}}:f'\left( e \right) = \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( e \right)}}{{x - e}} \cr
& {\text{we can note that }}f\left( x \right) = \ln x{\text{ and }}a = e \cr
& {\text{then}} \cr
& \,\,\,\,\,\,\,\,f'\left( x \right) = \frac{1}{x} \cr
& \,\,\,\,\,\,\,\,f'\left( e \right) = \frac{1}{e} \cr
& {\text{and}} \cr
& \,\,\,\,\,\,\,\mathop {\lim }\limits_{x \to e} \frac{{\ln x - 1}}{{x - e}} = f'\left( e \right) = \frac{1}{e} \cr} $$
