Answer
$$1$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 2} \frac{{\ln \left( {x - 1} \right)}}{{x - 2}} \cr
& {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr
& {\text{evaluate the limit substituting }}2{\text{ for }}x{\text{ into the numerator and }} \cr
& {\text{denominator}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 2} \ln \left( {x - 1} \right) = \ln \left( {2 - 1} \right) = 0 \cr
& \mathop {\lim }\limits_{x \to 2} \left( {x - 2} \right) = 2 - 2 = 0 \cr
& \cr
& {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr
& {\text{l'Hospital's rule applies}}{\text{. Differentiating the numerator and}} \cr
& {\text{denominator}}{\text{.}} \cr
& {\text{for ln}}\left( {x - 1} \right) \to {D_x}\left( {\ln \left( {x - 1} \right)} \right) = \frac{1}{{x - 1}} \cr
& {\text{for }}x - 2 \to {D_x}\left( {x - 2} \right) = 1 \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{\ln \left( {x - 1} \right)}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{\frac{1}{{x - 1}}}}{1} = \mathop {\lim }\limits_{x \to 2} \frac{1}{{x - 1}} \cr
& {\text{Find the limit of the quotient of the derivatives}} \cr
& = \frac{1}{{2 - 1}} \cr
& = 1 \cr
& {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr
& \mathop {\lim }\limits_{x \to 2} \frac{{\ln \left( {x - 1} \right)}}{{x - 2}} = 1 \cr} $$
