lim(x→∞)[（x^3+x^2+x^1+1)^(1/3) - x] 的极限。需要详细步骤。谢谢！

Answer 1

=1/3
A=（x^3+x^2+x^1+1)^(1/3) B= x首先分析分数同时乘以(A^2+B^2+AB)
然后上下同时除以X^2 然后比如x^2/x^3=0 之类的 结果是1/3

Answer 2

用等价无穷小代换
lim(x→∞)[（x^3+x^2+x^1+1)^(1/3) - x] 　（上下同除x）
=lim(x→∞)[（1+1/x+1/x^2+1/x^3)^(1/3) - 1] /(1/x) (1/x=t,x→∞,t→0）
=lim(t→0)[（1+t+t^2+t^3)^(1/3) - 1] /t (用等价无穷小代换）
=lim(t→0)(1/3t)/t
=1/3

Answer 3

lim(x→∞)[(x^3+x^2+x^1+1)^(1/3) - x]
=lim(x→∞)[(x^3+x^2+x^1+1)^(1/3) - x][(x^3+x^2+x^1+1)^(2/3)+ (x^3+x^2+x^1+1)^(1/3)x+x^3]/[(x^3+x^2+x^1+1)^(2/3)+ (x^3+x^2+x^1+1)^(1/3)x+x^3]
= lim(x→∞)(x^2+x^1+1)/[(x^3+x^2+x^1+1)^(2/3)+ (x^3+x^2+x^1+1)^(1/3)x+x^3]
= lim(x→∞)(1+1/x+1/x^2)/[(1+1/x+1/x^2+1/x^3)^(2/3)+(1+1/x+1/x^2+1/x^3)^(1/3)+1]
=1/3