高等数学问题求解

Question

如题所述

Answer 1

lim(x→-1/e-)y=+∞→铅锤渐进线x=-1/e  （不合题意）

斜渐近线：

k=lim(x→∞)y/x=lim(x→∞)ln(e+1/x)=lne=1

b=lim(x→∞)(y-kx)=lim(x→∞)[xln(e+1/x)-x]

=lim(1/x→0)[ln(e+1/x)-1]/(1/x)

=lim(1/x→0)[(1/x)'/(e+1/x)]/(1/x)'  洛必达法则

=lim(1/x→0)[1/(e+1/x)]

=1/e

斜渐近线：y=x+1/e

Answer 2

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