考研高等数学极限问题

2025-05-14 08:58:00
推荐回答(2个)
回答(1):

因为这里书写不便,故将我的答案做成图像贴于下方,谨供楼主参考(若图像显示过小,点击图片可放大)

回答(2):

(1+x)(1+x^2)...(1+x^(2^n))
=(1-x)(1+x)(1+x^2)...(1+x^(2^n)/(1-x)
=(1-x^2)(1+x^2)...(1+x^(2^n))/(1-x)
=...
=(1-x^(2^n))(1+x^(2^n))/(1-x)
=[1-x^(2^n)*x^(2^n)]/(1-x)
=[1-x^(2^(n+1))]/(1-x)
因为|x|<1,当n趋于∞时,x^(2^(n+1))趋于0
故原式的极限为:[1-0]/(1-x)=1/(1-x)
亲,不懂欢迎追问哦