\( \lim _ { x \rightarrow + } ( \frac { x } { x - 1 } - \frac { 1 } { \ln x } ) \)

To solve the limit \( \lim _ { x \rightarrow +\infty } \left( \frac { x } { x - 1 } - \frac { 1 } { \ln x } \right) \), let's analyze each term separately as \( x \) approaches infinity. First, consider the term \( \frac{x}{x - 1} \): \[ \frac{x}{x - 1} = \frac{x}{x(1 - \frac{1}{x})} = \frac{1}{1 - \frac{1}{x}}. \] As \( x \rightarrow +\infty \), \( \frac{1}{x} \rightarrow 0 \), so: \[ \frac{x}{x - 1} \rightarrow \frac{1}{1 - 0} = 1. \] Next, consider the term \( \frac{1}{\ln x} \): As \( x \rightarrow +\infty \), \( \ln x \) also approaches infinity, thus: \[ \frac{1}{\ln x} \rightarrow 0. \] Now, putting it all together, the limit simplifies to: \[ \lim _ { x \rightarrow +\infty } \left( \frac{x}{x - 1} - \frac{1}{\ln x} \right) = 1 - 0 = 1. \] Therefore, the final answer is: \[ \lim _ { x \rightarrow +\infty } \left( \frac { x } { x - 1 } - \frac { 1 } { \ln x } \right) = 1. \]