y=[e^(1/x^2)arctan(x^2+x+1)]/[(x-1)(x+2)]的渐近线有几条⊙﹏⊙b汗是y=[e^(1/x^2)]*arctan[(x^2+x+1)/[(x-1)(x+2)]]的渐近线有几条
y=[e^(1/x^2)arctan(x^2+x+1)]/[(x-1)(x+2)]的渐近线有几条
⊙﹏⊙b汗是
y=[e^(1/x^2)]*arctan[(x^2+x+1)/[(x-1)(x+2)]]的渐近线有几条
x->0时,y->-无穷
有一条渐近线x=0
x->无穷时,
limy=(1*π/2)/无穷=0
x->负无穷时
limy=(-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)
limx->无穷y'
=0+0-0-0
=0
limx->-无穷y'
=0
所以没有斜向的渐近线
渐近线x轴y轴
恩呢、这个好像是这样的可是y=[e^(x^(1/2)]*arctan[(x^2+x+1)/[(x-1)(x+2)]]的渐近线有几条郁闷这题目，。。，。
x=1和x=-2我忘了