Question

求第3题的极限和做法。。急！！！

Answer 1

解：∵(1-x^(1/3))(1+x^(1/3)+x^(2/3))=1-x
(1-x^(1/4))(1+x^(1/4))=1-x^(1/2)
(1-x^(1/2))(1+x^(1/2))=1-x
∴1-x^(1/4)=(1-x^(1/2))/(1+x^(1/4)).........(1)
1-x^(1/3)=(1-x)/(1+x^(1/3)+x^(2/3))..........(2)
1-x^(1/2)=(1-x)/(1+x^(1/2)).........(3)
故 原式=lim(x->1){(1-x^(1/2))[(1-x)/(1+x^(1/3)+x^(2/3))][(1-x^(1/2))/(1+x^(1/4))]/(1-x)³}
(由(1)和(2)代入)
=lim(x->1){(1-x)(1-x^(1/2))²/[(1-x)³(1+x^(1/3)+x^(2/3))(1+x^(1/4))]}
=lim(x->1){(1-x^(1/2))²/[(1-x)²(1+x^(1/3)+x^(2/3))(1+x^(1/4))]} (分子分母同除1-x)
=lim(x->1){[(1-x)/(1+x^(1/2))]²/[(1-x)²(1+x^(1/3)+x^(2/3))(1+x^(1/4))]} (由(3)代入)
=lim(x->1){(1-x)²/[(1-x)²(1+x^(1/2))²(1+x^(1/3)+x^(2/3))(1+x^(1/4))]}
=lim(x->1){1/[(1+x^(1/2))²(1+x^(1/3)+x^(2/3))(1+x^(1/4))]} (分子分母同除(1-x)²)
=1/[(1+1^(1/2))²(1+1^(1/3)+1^(2/3))(1+1^(1/4))]
=1/(2²*3*2)
=1/24。

Answer 2

看到你下面有积分的题目 应该学过等价无穷小吧 ，里面应该有关于1-更号x等价于1/2x的。另几个也类似

追问

但是等价无穷小的适用范围是x趋向0时，这里的x是趋向1！！！

追答

1减去更号x 就趋于0了 应该有的 不然你把(1-x)^3看做（1-更号x）^3（1+更号x）^3再和分子约掉，再用罗贝塔法则。

追问

那最后答案到底是什么？