求y=x^3⼀(x-1)^2的斜渐近线

2025-05-17 12:19:19
推荐回答(2个)
回答1:

定义域x-1≠0,x≠1
垂直渐近线为x=1

∵lim(x-->∞)[y-(x+2)]
=lim(x-->∞)[x³-(x-1)²(x+2)]/(x-1)
=lim(x-->∞)(3x-2)/(x²-2x+1)
=lim(x-->∞)(3/x-2/x²)/(1-2/x+1/x²)
=(0-0)/(1-0+0)
=0
又y-(x+2)≠0恒成立

∴曲线斜渐近线为y=x+2

回答2:

见图