Question

求y=xe^(2/x)+1的斜渐近线

Answer 1

k=lim(x->∞)y/x=lim(x->∞)[e^(2/x)+1/x]=1

b=lim)x->∞)(y-kx)=lim(x->∞)[xe^(2/x)+1-x]=lim(x->∞)=1+lim[e^(2/x)-1]/(1/x)=1+lim [-2/x^2*e^(2/x)/(-1/x^2)]=1+lim2e^(2/x)=1+2=3

故斜渐近线为y=x+3