曲线 $y=x \ln \left(e+\frac{1}{x-1}\right)$ 的渐近线方程为( )
(A) $y=x+e$
(B) $y=x+\frac{1}{e}$
(C) $y=x$
(D) $y=x-\frac{1}{e}$
$k=\lim _{x \rightarrow \infty} \frac{y}{x}=\lim _{x \rightarrow \infty} \frac{x \ln \left(e+\frac{1}{x-1}\right)}{x}=\lim _{x \rightarrow \infty} \ln \left(e+\frac{1}{x-1}\right)=1$,
$$
\begin{aligned}
b & =\lim _{x \rightarrow \infty}(y-k x)=\lim _{x \rightarrow \infty}\left[x \ln \left(e+\frac{1}{x-1}\right)-x\right]=\lim _{x \rightarrow \infty} x\left[\ln \left(e+\frac{1}{x-1}\right)-1\right] \\
& =\lim _{x \rightarrow \infty} x \ln \left[1+\frac{1}{e(x-1)}\right]=\lim _{x \rightarrow \infty} \frac{x}{e(x-1)}=\frac{1}{e}
\end{aligned}
$$

所以斜渐近线方程为 $y=x+\frac{1}{e}$.

因此答案为：B