曲线y=xln（e+1x）（x＞0）的渐近线方程为y=x+1ey=x+1e．

Question

曲线y=xln（e+(1)/(x)）（x＞0）的渐近线方程为

y=x+(1)/(e)

y=x+(1)/(e)

．

Answer 1

该曲线不存在垂直渐近线，且

lim

x→+∞

xln(e+(1)/(x))＝+∞，曲线不存在水平渐近线
a=

lim

x→+∞

(xln(e+(1)/(x)))/(x)＝

lim

x→+∞

ln(e+(1)/(x))＝1
b=

lim

x→+∞

(y−ax)＝

lim

x→+∞

x[ln(e+(1)/(x))−1]
=

lim

x→+∞

(ln(e+(1)/(x))−1)/((1)/(x))＝

lim

t→0+

(ln(e+t)−1)/(t)＝

lim

t→0+

(1)/(e+t)=(1)/(e)
所以渐近线方程为y=x+(1)/(e)
故答案为：y=x+(1)/(e)