lim[(ax+b)*-2]/(x-1)=1 X趨近於1,求a=?b=?
(ax+b)*=(ax+b)的1/3次方
2006-07-05 21:10:48 · 2 個解答 · 發問者 Tina Chang 1 in 教育與參考 ➔ 考試
lim ((ax+b)1/3- 2)/(x-1) = 1且 lim (x-1)= 0x→1 x→1∴lim (ax+b)1/3- 2 = 0 x→1(a+b)1/3- 2 = 0(a+b)1/3 = 2又 lim (a/3)(ax+b)-2/3/1 = 1 (羅必達法則)x→1(a/3)(a+b)-2/3 = 1a(a+b)-2/3 = 3a((a+b)1/3)-2 = 3a(2-2) = 3a = 12(12+b)1/3 = 212+b = 8b = - 4
2006-07-08 20:02:30 · answer #1 · answered by chan 5 · 0⤊ 0⤋
X趨近於1代入得lim(a+b)*/ ->0(趨近於0)=1
因為分母趨近於0
所以分子也要趨近於0
分子分母相消才會=1
所以(a+b)*-2=0
(a+b)*=2
所以a=1 b=1
代入下面的式子也成立
2006-07-05 21:53:05 · answer #2 · answered by 家良 3 · 0⤊ 0⤋