求lim x->n [x-[x-1]]的詳細解法 謝謝?
回答 (1)
2017-11-05 5:59 pm
求lim x->n [x-[x-1]]的詳細解法 謝謝?
Sol
n為整數
lim x->n+_[x－1]
=lim x->n_[n+0.001－1]
=lim x->n_[n－0.999]
=n-1
lim x->n+_[x－[x－1]]
=lim x->n+_[x－(n－1)]
=lim x->n_[n+0.001－n+1]
=lim x->n_[1.001]
=1
lim x->n-_[x－1]
lim x->n__[n－0.001－1]
=lim x->n_[n－1.001]
=n－2
lim x->n-_[x－[x－1]]
=lim x->n-_[x－(n－2)]
=lim x->n_[n-0.001－n+2]
=lim x->n_[1.999]
=1
So
lim x->n__[x－n+2]
=1
Sol
n為整數
lim x->n+_[x－1]
=lim x->n_[n+0.001－1]
=lim x->n_[n－0.999]
=n-1
lim x->n+_[x－[x－1]]
=lim x->n+_[x－(n－1)]
=lim x->n_[n+0.001－n+1]
=lim x->n_[1.001]
=1
lim x->n-_[x－1]
lim x->n__[n－0.001－1]
=lim x->n_[n－1.001]
=n－2
lim x->n-_[x－[x－1]]
=lim x->n-_[x－(n－2)]
=lim x->n_[n-0.001－n+2]
=lim x->n_[1.999]
=1
So
lim x->n__[x－n+2]
=1
收錄日期: 2021-04-30 22:37:05
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20171105074404AAezsso