求y=(2x^2-3x+3)/(x-1)渐近线
人气:227 ℃ 时间:2020-05-21 00:16:28
解答
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.
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