Evaulate this:
(lim h-->0) [square root(9+h) - 3] / h
thx!
Math~!
2007-11-23 6:51 pm
回答 (2)
2007-11-23 7:16 pm
✔ 最佳答案
(lim h-->0) [square root(9+h) - 3] / h----> [square root(9+h) - 3]*[square root(9+h) + 3]/ h*[square root(9+h) + 3]
----> (9+h) -9/h*square root(9+h) + 3h
----> h/h*square root(9+h) + 3h
----> 1/square root(9+h)+3
(lim h-->0)
----> 1/square root(9)+3
----> 1/6
2007-11-23 7:18 pm
lim h-->0 [(9+h)^0.5 - 3]/h
lim h-->0 [(9+h)^0.5 - 3][(9+h)^0.5 + 3] / h[(9+h)^0.5 + 3]
lim h-->0 [(9+h) - 9] / h[(9+h)^0.5 + 3]
lim h-->0 h / h[(9+h)^0.5 + 3
lim h-->0 1 / (9+h)^0.5 + 3
h does not equal to 0
hence the limit is 1 / (9+0)^0.5 + 3 = 1/6
lim h-->0 [(9+h)^0.5 - 3][(9+h)^0.5 + 3] / h[(9+h)^0.5 + 3]
lim h-->0 [(9+h) - 9] / h[(9+h)^0.5 + 3]
lim h-->0 h / h[(9+h)^0.5 + 3
lim h-->0 1 / (9+h)^0.5 + 3
h does not equal to 0
hence the limit is 1 / (9+0)^0.5 + 3 = 1/6
參考: me
收錄日期: 2021-04-14 00:52:43
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071123000051KK00847