Answer
$$ - \frac{1}{e}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to e} \frac{{1 - \ln x}}{{\left( {x - e} \right)\ln x}} \cr
& = \mathop {\lim }\limits_{x \to e} \frac{{\frac{1}{{\ln x}} - \frac{{\ln x}}{{\ln x}}}}{{\frac{{\left( {x - e} \right)\ln x}}{{\ln x}}}} \cr
& = \mathop {\lim }\limits_{x \to e} \frac{{\frac{1}{{\ln x}} - 1}}{{x - e}} = \mathop {\lim }\limits_{x \to e} \frac{{\frac{1}{{\ln x}} - \frac{1}{{\ln \left( e \right)}}}}{{x - e}} \cr
& {\text{From the definition of the derivative }} \cr
& f'\left( c \right) = \mathop {\lim }\limits_{x \to c} \frac{{f\left( x \right) - f\left( c \right)}}{{x - c}} \cr
& \underbrace {\mathop {\lim }\limits_{x \to e} \frac{{\frac{1}{{\ln x}} - \frac{1}{{\ln \left( e \right)}}}}{{x - e}}}_{f'\left( c \right) = \mathop {\lim }\limits_{x \to c} \frac{{f\left( x \right) - f\left( c \right)}}{{x - c}}} \Rightarrow f\left( x \right) = \frac{1}{{\ln x}},{\text{ }}a = 3 \cr
& {\text{Then }} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{{\ln x}}} \right] = - {\left( {\ln x} \right)^{ - 2}}\left( {\frac{1}{x}} \right) \cr
& f'\left( x \right) = - \frac{1}{{x{{\left( {\ln x} \right)}^2}}} \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to e} \frac{{1 - \ln x}}{{\left( {x - e} \right)\ln x}} = f'\left( e \right) \cr
& f'\left( e \right) = - \frac{1}{{e{{\left( {\ln e} \right)}^2}}} = - \frac{1}{e} \cr
& \mathop {\lim }\limits_{x \to e} \frac{{1 - \ln x}}{{\left( {x - e} \right)\ln x}} = - \frac{1}{e} \cr} $$
