求y=(2x^2-3x+3)/(x-1)渐近线

2个回答

  • 显然,当x趋向1时,y趋向∞,∴x=1是函数的竖直渐近线.

    令函数的斜渐近线为:y=ax+b,则:

    a=lim(x→∞){(2x^2-3x+3)/[x(x-1)]}

    =lim(x→∞)[(2-3/x+3/x^3)/(1-1/x)]=2,

    b=lim(x→∞)[(2x^2-3x+3)/(x-1)-2x]

    =lim(x→∞)[(2x^2-3x+3-2x^2+2x)/(x-1)]=lim(x→∞)[(-x+3)/(x-1)]

    =lim(x→∞)[-1+3/x)/(1-1/x)]=-1.

    ∴函数的斜渐近线是:y=2x-1.

    综上所述,得函数的渐近线有两条,分别是:x=1、y=2x-1.