曲线y=1-x+x4+1x2+1（）A．没有渐近线B．只有一条渐近线C．共有两条渐近线D．共有三条渐近线

Question

曲线y=1-x+√((x4+1)/(x2+1))（　　）

A．没有渐近线
B．只有一条渐近线
C．共有两条渐近线
D．共有三条渐近线

Answer 1

因为y（x）没有间断点，故只需计算曲线的斜渐近线即可．
（1）x→+∞时的情形：
因为

lim

x→+∞

(y(x))/(x)=

lim

x→+∞

((1)/(x)−1+(1)/(x)√((x4+1)/(x2+1)))
=

lim

x→+∞

((1)/(x)−1+√((x4+1)/(x4+x2)))
=0，

lim

x→+∞

(1−x+√((x4+1)/(x2+1)))=

lim

x→+∞

(1+(√(x4+1)−x√(x2+1))/(√(x2+1)))
=

lim

x→+∞

(1+(1−x2)/(√(x2+1)(√(x4+1)+x√(x2+1))))
=

lim

x→−∞

(1+((1)/(x3)−(1)/(x))/(√(1+(1)/(x2))(√(1+(1)/(x4))+√(1+(1)/(x2))))) 
=1，
所以，当x→+∞时，曲线存在渐近线y=1．
（2）x→-∞时的情形：
因为

lim

x→−∞

(y(x))/(x)=

lim

x→−∞

((1)/(x)−1+(1)/(x)√((x4+1)/(x2+1)))
=

lim

x→−∞

((1)/(x)−1−√((x4+1)/(x4+x2)))
=-2，

lim

x→−∞

(y(x)+2x)=

lim

x→−∞

(1+x+√((x4+1)/(x2+1)))
=

lim

x→−∞

(1+(x√(x2+1)+√(x4+1))/(√(x2+1)))
=

lim

x→−∞

(1+(x2−1)/(√(x2+1)(x√(x2+1)−√(x4+1))))
=

lim

x→−∞

(1+((1)/(x)−(1)/(x3))/(√(1+(1)/(x2))(√(1+(1)/(x2))+√(1+(1)/(x4)))))
=1，
所以，当x→+∞时，曲线存在渐近线y=-2x+1．
综上，曲线y（x）存在两条渐近线．
故选：C．