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(x的m次方-1)/(x的n次方-1),当X趋于1时的极限是?
人气:484 ℃ 时间:2019-08-20 04:07:57
解答
x^m-1
=(x-1)[x^(m-1)+x^(m-2)+……+x+1]
x^n-1
=(x-1)[x^(n-1)+x^(n-2)+……+x+1]
所以原式=[x^(m-1)+x^(m-2)+……+x+1]/[x^(n-1)+x^(n-2)+……+x+1]
极限=(1+1+……+1)/(1+1+……+1)=m/n
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