#lim_(x to 1) (1/ln(x)-1/(x-1))=lim_(x to 1) (x-1-ln(x))/(ln(x)(x-1))=[0/0]#
And now to get rid of #0/0# you can use the de L'Hôspital's Rule which states that when evaluating #0/0# or #infty/infty# indeterminate forms the limit of the quotient stays the same if derivatives of the numerator and denominator (evaluated seperately, not using the formula for the derivative of a quotient) are substituted for the numerator and denominator, respectively (as long as the 'new' limit exists).
So (in this case using the de L'Hôpital's Rule twice):
#lim_(x to 1) (1-1/x)/((x-1)/x +ln(x))=[0/0]=lim_(x to 1) (1/x^2)/(1/x^2 +1/x)=[1/(1+1)]=1/2#