Answer
$\lim\limits_{x \to 1^+} (\frac{x}{x-1}-\frac{1}{ln~x}) = \frac{1}{2}$
Work Step by Step
$\lim\limits_{x \to 1^+} (\frac{x}{x-1}-\frac{1}{ln~x})$
$= \lim\limits_{x \to 1^+} [\frac{x~ln~x}{(x-1)~ln~x}-\frac{x-1}{(x-1)~ln~x}]$
$= \lim\limits_{x \to 1^+} [\frac{x~ln~x-x+1}{(x-1)~ln~x}] = \frac{0}{0}$
We can apply L'Hospital's Rule:
$\lim\limits_{x \to 1^+} [\frac{ln~x+x~(1/x)-1}{ln~x+(x-1)/x}]$
$= \lim\limits_{x \to 1^+} [\frac{ln~x+1-1}{ln~x+(x-1)/x}] = \frac{0}{0}$
We can apply L'Hospital's Rule:
$= \lim\limits_{x \to 1^+} \frac{1/x}{(1/x)+(1/x^2)} = \frac{1}{1+1} = \frac{1}{2}$
Therefore:
$\lim\limits_{x \to 1^+} (\frac{x}{x-1}-\frac{1}{ln~x}) = \frac{1}{2}$