1. ## L'hopital's rule

lim [((x)/(x-1))-((1)/(ln x))]
x->1+

use l'hopitals rule

2. Subtract the fractions to get an indeterminate form: $\lim_{x \to 1^+} \left( \frac{x}{x-1} - \frac{1}{\ln x}\right) = \lim_{x \to 1^+} \frac{{\color{red}x\ln x} - {\color{blue}x} + 1}{{\color{magenta}(x-1)\ln x}} = \left[\frac{0}{0}\right]$

Use L'hopital's rule to get (product rule is needed to take care of the xlnx): $= \lim_{x \to 1^+} \frac{ {\color{red}\ln x + 1} - {\color{blue}1}}{{\color{magenta}\ln x + 1 -\frac{1}{x}}} = \lim_{x \to 1^+} \frac{\ln x}{\ln x + 1 -\frac{1}{x}} = \left[\frac{0}{0}\right]$

This is another indeterminate form so apply L'hopital's rule again. Try taking the limit and if you get a definite answer then you're done.