Answer
$\frac{1}{2}.$
Work Step by Step
We have
$$
\lim _{x \rightarrow 1}\frac{x(\ln x-1)+1}{(x-1)\ln x}=\lim _{x \rightarrow 1}\frac{x(\ln x-1)+1}{(x-1)\ln x}=\frac{0}{0}
$$
which is an intermediate form, so we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow 1}\frac{x(\ln x-1)+1}{(x-1)\ln x}=\lim _{x \rightarrow 1}\frac{(\ln x-1)+x(1/x)}{\ln x+(x-1)(1/x)}\\
=\lim _{x \rightarrow 1}\frac{\ln x}{\ln x+1-(1/x)}=\frac{0}{0}
$$
which is an intermediate form, so we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow 1}\frac{\ln x}{\ln x+1-(1/x)}=\lim _{x \rightarrow 1}\frac{1/ x}{1/x+(1/x^2)}=\frac{0}{0}
=\frac{1}{2}.
$$
