设斜渐近线为y=ax+ba=lim[x→∞] y/x=lim[x→∞] ln(e+1/x)=1b=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...