当x趋向于正无穷,求三次根号(x^3+x^2+x+1)-x的极限

Answer 1

lim三次根号(x^3+x^2+x+1)-x (令 x=1/t)
=lim三次根号(1/t^3+1/t^2+1/t+1)-1/t
=lim [三次根号(1+t+t^2+t^3)-1]/t
=lim(t+t^2+t^3)/3t
=lim(1+t+t^2)/3 (t趋向于0+)
=1/3

Answer 2

lim(x->+∞) [ ³√(x^3+x^2+x+1) - x ]
= lim(x->+∞) x * [ ³√(1+1/x +1/x² +1/x³) - 1 ] 令t=1/x， t->0
= lim(t->0+) [ ³√(1+t +t² +t³) - 1 ] / t 等价无穷小代换 ³√(1+t +t² +t³) - 1 ~ t /3
= lim(t->0+) (t/3) /t
= 1/3

Answer 3

√(x^2+x+1)-√(x^2-x+1)
分子分母同时乘√(x²+x+1)+√(x²-x+1)
=[√(x^2+x+1)-√(x^2-x+1)][[√(x^2+x+1)+√(x^2-x+1)]]/[√(x^2+x+1)+√(x^2-x+1)]
=2x/[√(x^2+x+1)+√(x^2-x+1)]
=2/[√(1+1/x+1/x^2)+√(1-1/x+1/x^2)]
=2/(1+1)
=1