Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 13

lim_(x->1^+)(x/(x-1)-1/ln(x))
=lim_(x->1^+)(xln(x)-1(x-1))/((x-1)ln(x))
=lim_(x->1^+)(xln(x)-x+1)/((x-1)ln(x))
Apply L'Hospital's rule, Test L'Hospital condition :0/0
=lim_(x->1^+)((xln(x)-x+1)')/(((x-1)ln(x))')
=lim_(x->1^+)(x(1/x)+ln(x)-1)/((x-1)(1/x)+ln(x))
=lim_(x->1^+)ln(x)/((x-1)/x+ln(x))
=lim_(x->1^+)(xln(x))/(x-1+xln(x))
Again apply L'Hospital's rule, Test L'Hospital condition:0/0
=lim_(x->1^+)((xln(x))')/((x-1+xln(x))')
=lim_(x->1^+)(x(1/x)+ln(x))/(1+x(1/x)+ln(x))
=lim_(x->1^+)(1+ln(x))/(2+ln(x))
Now plug in the value and simplify,
=(1+ln(1))/(2+ln(1))
=1/2