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
$$\lim _{x \rightarrow 1} \frac{1}{\ln x} \rightarrow \infty ,$$
so the limit is indeterminate. Here we use a common denominator:
$$\begin{aligned}
\lim _{x \rightarrow 1}\left(\frac{x}{x-1}-\frac{1}{\ln x}\right) &=\lim _{x \rightarrow 1} \frac{x \ln x-(x-1)}{(x-1) \ln x} \rightarrow \frac{0}{0} \\
&\stackrel{\text { H }}{=} \lim _{x \rightarrow 1} \frac{x(1 / x)+\ln x-1}{(x-1)(1 / x)+\ln x} \\
&=\lim _{x \rightarrow 1} \frac{\ln x}{1-(1 / x)+\ln x} \\
&=\lim _{x \rightarrow 1} \frac{1 / x}{1 / x^{2}+1 / x} \cdot \frac{x^{2}}{x^{2}} \\
&=\lim _{x \rightarrow 1} \frac{x}{1+x} \\
&=\frac{1}{1+1}\\
&=\frac{1}{2}.
\end{aligned}
$$
Note that the use of l’Hospital’s Rule is justified because $ x \ln x-(x-1) \rightarrow 0 $ and $(x-1) \ln x \rightarrow 0 $ as$ \rightarrow 1$
