#lim_(x to oo) (xln(x+1))/(x-1)# gives #oo/oo# so we can use L'Hopital's rule:
#lim_(x to oo) (xln(x+1))/(x-1) = lim_(x to oo) (d/dx(xln(x+1)))/(d/dx((x-1))#
#=lim_(x to oo) (xln(x+1))/(x-1)=lim_(x to oo) ((x)/(x+1)+ln(x+1))/1 #
which goes to #oo# as #x to oo#.
But maybe the question is:
#lim_(x to oo) (x*ln((x+1)/(x-1)))#
that goes to #oo*0# so we need to manipulate it before using L'Hopital's Rule:
#lim_(x to oo) ((ln((x+1)/(x-1)))/(1/x))#
is algebraically equivalent but goes to #0/0# as #x to oo# so we can use L'Hopital's Rule on it.
#lim_(x to oo) ((ln((x+1)/(x-1)))/(1/x)) = lim_(x to oo) ((d/dx(ln((x+1)/(x-1))))/(d/dx(1/x)))#
#=lim_(x to oo) (((x-1)/(x+1)*((x-1)(1)-(x+1)(1))/(x-1)^2))/(-1/x^2)#
#=lim_(x to oo) (((x-1)/(x+1)*((-2)/(x-1)^2))/(-1/x^2))#
#=lim_(x to oo) ((((-2)/(x^2-1)))/(-1/x^2))#
#=lim_(x to oo) ((2x^2)/(x^2-1)) = 2#
