lim {x->+} (ln (x+1) / (x-1))

= lim {x->+} ((ln x+1)/(x+1) (x+1)/(x-1)) = 0

On reconnait la forme lim {X->+} (ln X / X) = 0
et on voit que (x+1)(x-1) = (x (1+1/x) (x (1-x) ) --> 1