lim(x→∞)( 開方(x+1) - 開方(x-1) )
請問答案係咩野
詳細步驟 please
求極限lim(x→∞) 好簡單 請入來睇睇
2011-05-17 5:28 am
回答 (3)
2011-05-17 5:37 am
✔ 最佳答案
lim(x→∞) (√(x+1) - √(x-1))= lim(x→∞) (√(x+1) - √(x-1)) (√(x+1) + √(x-1)) / (√(x+1) + √(x-1))= lim(x→∞) ((x+1) - (x-1)) / (√(x+1) + √(x-1))
= lim(x→∞) 2 / (√(x+1) + √(x-1))
= lim(x→∞) (2/√x) / [(√(x+1) + √(x-1)) / √x]
= lim(x→∞) (2/√x) / (√(1+1/x) + √(1-1/x))
= 0 / (1 + 1)
= 0
2011-05-18 5:13 am
lim(x→∞)( √(x+1) - √(x-1) )
= lim(x→∞)( √(x+1) - √(x-1) )( √(x+1) + √(x-1) )/( √(x+1) + √(x-1) )
= lim(x→∞) (x + 1 - x + 1)/( √(x+1) + √(x-1) )
= lim(x→∞) 2/( √(x+1) + √(x-1) )
= 0
= lim(x→∞)( √(x+1) - √(x-1) )( √(x+1) + √(x-1) )/( √(x+1) + √(x-1) )
= lim(x→∞) (x + 1 - x + 1)/( √(x+1) + √(x-1) )
= lim(x→∞) 2/( √(x+1) + √(x-1) )
= 0
2011-05-17 5:37 am
lim(x→∞)( √(x+1) - √(x-1) )
= lim(x→∞)( √(x+1) - √(x-1) )( √(x+1) + √(x-1) )/( √(x+1) + √(x-1) )
= lim(x→∞) (x + 1 - x + 1)/( √(x+1) + √(x-1) )
= lim(x→∞) 2/( √(x+1) + √(x-1) )
= 0
= lim(x→∞)( √(x+1) - √(x-1) )( √(x+1) + √(x-1) )/( √(x+1) + √(x-1) )
= lim(x→∞) (x + 1 - x + 1)/( √(x+1) + √(x-1) )
= lim(x→∞) 2/( √(x+1) + √(x-1) )
= 0
收錄日期: 2021-04-21 22:23:58
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110516000051KK01121