Question

曲线y=xln(e+1/x)(x>0)的斜渐近线方程为（求详细点）

Answer 1

设斜渐近线为y=ax+b
a=lim[x→∞] y/x=lim[x→∞] ln(e+1/x)=1
b=lim[x→∞] [xln(e+1/x)-ax]
=lim[x→∞] [xln(e+1/x)-x]
=lim[x→∞] [xln(e+1/x)-xlne]
=lim[x→∞] xln[(e+1/x)/e]
=lim[x→∞] xln[1+1/(ex)]

等价无穷小代换
=lim[x→∞] x/(ex)
=1/e

因此渐近线为：y=x + 1/e

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Answer 2

斜渐近线的斜率为：
k = lim {x->无穷大} y/x
= lim {x->无穷大} ln(e+1/x)
= 1
再看lim {x->无穷大} (y-kx)
=lim {x->无穷大} xln(e+1/x) - x
=lim {x->无穷大} x [ln(e+1/x) - 1]
=lim {x->无穷大} x {ln[(e+1/x)/e]}
=lim {x->无穷大} x ln(1+1/ex)
令1/x=t
=lim {t->0} [ln(1+t/e)] / t
=1/e
所以斜渐近线是y=x+1/e