$\displaystyle \lim_{x\to 1} \dfrac {(x^x - x)}{x-1 - \log x} \left [ \dfrac {0}{0} \right ] $ $ \displaystyle = \lim_{x\to 1} \dfrac {x^x (1+ \log x)-1} {1- \frac {1}{x}} \left [ \dfrac {0}{0} \right ]$ $\displaystyle = \lim_{x\to 1} \dfrac {x^x (1+ \log x)^2 + x^x \left ( \frac {1}{x} \right )}{1- \frac {1}{x}}$ $= \dfrac {1+1}{1}$ $= 2$ $\displaystyle \lim_{x\to 1} \dfrac {(x^x - x)}{x-1-\log x} = 2$