求³√(x³-5x²+1)-x的极限,x趋于无穷大
2个回答

为了表示方便,令y=3√(x^3-5x^2+1) 所以:y^3=x^3-5x^2+1

以下为根限运算:

3√(x^3-5x^2+1)-x=y-x=(y^3-x^3)/(y^2-xy+x^2)立方差公式

=(-5x^2+1)/(y^2-xy+x^2) (将y^3=x^3-5x^2+1代入)

=-5x^2/(y^2-xy+x^2)+1/(y^2-xy+x^2)后一项根限为0

=-5x^2/(y^2-xy+x^2)因为y/x的根限为1

=-5/[(y/x)^2-(y/x)+1]=-5/(1-1+1)=-5