Answer
$$
\lim _{x \rightarrow 1^{+}}\left(\frac{x}{x-1}-\frac{1}{\ln x}\right) =\frac{1}{2}
$$
Work Step by Step
$$
\lim _{x \rightarrow 1^{+}}\left(\frac{x}{x-1}-\frac{1}{\ln x}\right)
$$
First notice that
$$
\lim _{x \rightarrow 1^{+}} \frac{x}{x-1} \,\,\, \rightarrow \infty
$$
and
$$
\lim _{x \rightarrow 0^{+}} \frac{1}{\ln x}\,\,\, \rightarrow \infty ,
$$
so the limit is indeterminate and has the form $\infty -\infty $. Here we use a common denominator and l’Hospital’s Rule is justified, then we gives:
$$\begin{aligned}
\lim _{x \rightarrow 1^{+}}\left(\frac{x}{x-1}-\frac{1}{\ln x}\right) &=\lim _{x \rightarrow 1^{+}}\left(\frac{x \ln x-x+1}{(x-1) \ln x}\right) \rightarrow \frac{0}{0} \\
& \stackrel{\mathrm{H}}{=} \lim _{x \rightarrow 1^{+}} \frac{x \cdot(1 / x)+\ln x-1}{(x-1) \cdot(1 / x)+\ln x} \\
&=\lim _{x \rightarrow 1^{+}} \frac{\ln x}{1-1 / x+\ln x} \rightarrow \frac{0}{0} \\
& \stackrel{\mathrm{H}}{=} \lim _{x \rightarrow 1^{+}} \frac{1 / x}{1 / x^{2}+1 / x} \\
&=\frac{1}{1+1} \\
&=\frac{1}{2}.
\end{aligned}$$
