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