Question

y=[e^(1/x^2)arctan(x^2+x+1)]/[(x-1)(x+2)] 的渐近线有几条

Answer 1

x->0时,y->-无穷

有一条渐近线 x=0

x->无穷时,

lim y=(1*π/2)/无穷=0

x->负无穷时

lim y=(-π/2)/(-无穷)=0

所以有渐近线 y=0

y'=(-2/x³)e^(1/x²)arctan(x²+x+1)(x-1)^(-1)(x+2)^(-1)

+e^(1/x²)[(2x+1)/((x²+x+1)²+1)](x-1)^(-1)(x+2)^(-1)

-e^(1/x²)arctan(x²+x+1)(x-1)^(-2)(x+2)^(-1)

-e^(1/x²)arctan(x²+x+1)(x-1)^(-1)(x+2)^(-2)

lim x->无穷 y'

=0+0-0-0

=0

lim x->-无穷 y'

=0

所以没有斜向的渐近线

渐近线x轴y轴