y=(3x^2-1)/(x+1)e^（2/x）的斜渐近线

Question

y=(3x^2-1)/(x+1)e^（2/x）的斜渐近线

Answer 1

设渐近线方程：y=kx+b
则k=lim(x→∞)[(3x²-1)/(x-1)][e^(2/x)]/x
=lim(x→∞)[(3x²-1)e^(2/x)]/[x(x-1)]
=lim(x→∞)[(3-1/x²)e^(2/x)]/(1-1/x)
=[(3-0)e^0]/(1-0)
=3
故b=lim(x→∞)(y-kx)
=lim(x→∞)[(3x²-1)e^(2/x)/(x+1)]-3x
=lim(x→∞)[(3x²-1)e^(2/x)-3x²-3x]/(x+1)
=lim(x→∞)[3x²[e^(2/x)-1]-3x-e^(2/x)]/(x+1)
=lim(x→∞){3x[e^(2/x)-1]-3-[e^(2/x)]/x}/(1+1/x)
=lim(x→∞){3x·2/x-3-[e^(2/x)]/x}/(1+1/x)
=lim(x→∞){6-3-[e^(2/x)]/x}/(1+1/x)
=3
故渐近线方程：y=3x+3