lim(1+x)(1+x^2)…[1+x^(2^n)],|x|

来源：学生作业帮助网编辑：作业帮时间：2024/11/15 11:59:30

lim(1+x)(1+x^2)…[1+x^(2^n)],|x|
lim(1+x)(1+x^2)…[1+x^(2^n)],|x|

lim(1+x)(1+x^2)…[1+x^(2^n)],|x|
lim(1+x)(1+x^2)…[1+x^(2^n)]
=lim(1-x)(1+x)(1+x^2)…[1+x^(2^n)]/(1-x)
=lim(1-x^2)(1+x^2)…[1+x^(2^n)]/(1-x)
.逐项乘下去得：
=lim[1-x^(2^n+1)]/(1-x)
因为|x|<1,所以lim x^(2^n+1)=0
所以原式=lim1/(1-x),这里应该是n→∞,而不是x→∞
答案就是1/(1-x)

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