Answer
$$0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \ln \left( {\frac{{x + 1}}{{x - 1}}} \right) \cr
& {\text{Evaluate}} \cr
& \mathop {\lim }\limits_{x \to \infty } \ln \left( {\frac{{x + 1}}{{x - 1}}} \right) = \ln \left( {\frac{{\infty + 1}}{{\infty - 1}}} \right) = \infty \cr
& {\text{By the long division}}\frac{{x + 1}}{{x - 1}} = 1 + \frac{2}{{x - 1}},{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to \infty } \ln \left( {\frac{{x + 1}}{{x - 1}}} \right) = \mathop {\lim }\limits_{x \to \infty } \ln \left( {1 + \frac{2}{{x - 1}}} \right) \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{x \to \infty } \ln \left( {1 + \frac{2}{{x - 1}}} \right) = \ln \left( {1 + \frac{2}{{\infty - 1}}} \right) \cr
& = \ln \left( {1 + \frac{2}{\infty }} \right) \cr
& = \ln \left( 1 \right) \cr
& = 0 \cr} $$
