求lim(n->∞) [n-√ (n^2-1)]的值 謝謝~?
回答 (3)
2016-07-28 2:55 am
lim(n->∞) [n-√(n^2-1)]
=lim(n->∞) [n-√(n^2-1)][n+√(n^2-1)]/[n+√(n^2-1)]
=lim(n->∞) (n^2 -n^2 -1)/[n+√ (n^2-1)]
=lim(n->∞) 1/[n+√ (n^2-1)]
=0
因為分母的數字是非常之大，所以接近零
=lim(n->∞) [n-√(n^2-1)][n+√(n^2-1)]/[n+√(n^2-1)]
=lim(n->∞) (n^2 -n^2 -1)/[n+√ (n^2-1)]
=lim(n->∞) 1/[n+√ (n^2-1)]
=0
因為分母的數字是非常之大，所以接近零
2016-05-15 1:51 pm
n-sqrt(n^2-1)
=n-n*sqrt(1-1/n^2)
=n(1-sqrt(1-1/n^2)
=n(1-[1-(1/2)(1/n^2)^2-(1/8)(1/n^2)^4-(1/16)(1/n^2)^6-(1/128)(1/n^2)^8....])
=n(1-[1-0-0-0....]) n->inf
=0
=n-n*sqrt(1-1/n^2)
=n(1-sqrt(1-1/n^2)
=n(1-[1-(1/2)(1/n^2)^2-(1/8)(1/n^2)^4-(1/16)(1/n^2)^6-(1/128)(1/n^2)^8....])
=n(1-[1-0-0-0....]) n->inf
=0
2016-05-15 11:01 am
Sol
lim(n->∞)_[n－√(n^2－1)]
=lim(n->∞)_[n－√(n^2－1)]*[n+√(n^2－1)]/[n+√(n^2－1)]
=lim(n->∞)_[n^2－(n^2－1)]/[n+√(n^2－1)]
=lim(n->∞)_1/[n+√(n^2－1)]
=0
lim(n->∞)_[n－√(n^2－1)]
=lim(n->∞)_[n－√(n^2－1)]*[n+√(n^2－1)]/[n+√(n^2－1)]
=lim(n->∞)_[n^2－(n^2－1)]/[n+√(n^2－1)]
=lim(n->∞)_1/[n+√(n^2－1)]
=0
收錄日期: 2021-04-18 14:49:28
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160515025531AAaZJC6