(2023年·数学一、数学二·分值5分)曲线y=xln(e+(1)/(x-1))的斜渐近线方程为A. y=x+e.B. y=x+(1)/(e).C. y=x.D. y=x-(1)/(e).

本题考查曲线斜渐近线方程的求解，关键是计算斜渐近线的斜率$a$和截距$b$，公式为：若$\lim_{xx \to \infty} \frac{y}{x} = a$且$a \neq 0$，则$\lim_{x \to \infty} (y - ax) = b$，则斜渐近线方程为$y = ax + b$。

步骤1：计算斜率$a$

$a = \lim_{x \to \infty} \frac{y}{x} = \lim_{x \to \infty}frac{x\ln\left(e + \frac{1}{x - 1}\right)}{x} = \lim_{x \to \infty}\ln\left(e + \frac{1}{x - 1}\right) \quad\text{（约掉}\,x\text{）}$
当$x \to \infty$时，$\frac{1}{x - 1} \to 0$，故：
$a = \lnln(e + 0) = \ln e = 1$

步骤2：计算截距$b$

$b = \lim_{x \to \infty (y - ax) = \lim_{x \to \infty}\left[x\ln\left(e + \frac{1}{x - 1}\right) - x\right] = \lim_{x \to \infty}x\left[\ln\left(e + \frac{1}{x - 1}\right) - 1\right]$
化简对数：
$\ln\left(e + \frac{1}{}{x - 1}\right) - 1 = \ln\left(e\left(1 + \frac{1}{e(x - 1)}\right)\right) - 1 = \ln e + \ln\left(1 + \frac{1}{e(x - 1)}\right) - 1 = \ln\left(1 + \frac{1}{e(x - 1)}\right)$
当$t \to 0$时，$\ln(1 + t) \sim t$，令$t = \frac{1}{e(x - 1)}$，则$t \to 0$，故：
$\ln\left(1 + \frac{1}{e(x - 1)}\right) \sim \frac{1}{e(x - 1)}$
代入得：
$b = \lim_{x \to \infty}x \cdot \frac{1}{e(x - 1)} = \lim_{x \to \infty}\frac{x}{e(x - 1)} = \frac{1}{e}$

结论

斜渐近线方程为$y = x + \frac{1}{e}$，对应选项B。