$\displaystyle\lim_{x \to 1} \left(\dfrac{x}{x-1}-\dfrac{1}{\ln x}\right)\\ = \displaystyle\lim_{x \to 1} \dfrac{x\ln x-x+1}{(x-1)\ln x}\\ = \displaystyle\lim_{x \to 1} \dfrac{1+\ln x-1}{\ln x+\dfrac{x-1}{x}}(L'Hopital)\\ = \displaystyle\lim_{x \to 1} \dfrac{\ln x}{\ln x+\dfrac{x-1}{x}}\\ = \displaystyle\lim_{x \to 1} \dfrac{x\ln x}{x\ln x+x-1}\\ =\displaystyle\lim_{x \to 1} \dfrac{\ln x+1}{\ln x+1+1}(L'Hopital)\\ =\displaystyle\lim_{x \to 1} \dfrac{\ln 1+`1}{\ln 1+1+1}\\ =\dfrac{1}{2}$
