Re: [微分] 一題極限
※ 引述《ian5j6 (Persistent Effort)》之銘言:
: ※ 引述《ian5j6 (Persistent Effort)》之銘言:
: limit (x^2)sin(1/x)-x =?
: x->無窮大
: 請問可以有那位好心的大大有過程嗎?
: limit (x^2)[sin(1/x)-(1/x)] then??
多一步好了
lim (θ-sinθ)/θ^3 = 1/6
θ->0
lim (θ-sinθ) = lim (1/6)θ^3 = 0
θ->0 θ->0
-------------------------------------------
(虛線以上是你需要的,實際上可以由級數展開而得.)
所以
limit (x^2)[sin(1/x)-(1/x)] = lim x^2 * (-1/6) * (1/x^3) = lim (-1/6)*(1/x)
x->∞ x->∞ x->∞
= 0
---------------------------------------------
(再重寫一次,誰叫我很嫩.)
sin(t) - t
limit (x^2)[sin(1/x)-(1/x)] = lim ---------------- = lim (-1/6)*t = 0
x->∞ t->0 t^2 t->0
這樣大概是直接將"已知極限"用進去了吧.
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 118.233.32.58
→
12/23 23:58, , 1F
12/23 23:58, 1F
→
12/23 23:59, , 2F
12/23 23:59, 2F
→
12/24 00:00, , 3F
12/24 00:00, 3F
※ 編輯: zptdaniel 來自: 118.233.32.58 (12/24 00:18)
推
12/24 00:25, , 4F
12/24 00:25, 4F
推
12/24 00:42, , 5F
12/24 00:42, 5F
推
12/24 22:17, , 6F
12/24 22:17, 6F
→
12/24 22:18, , 7F
12/24 22:18, 7F
→
12/24 23:01, , 8F
12/24 23:01, 8F
※ 編輯: zptdaniel 來自: 118.233.32.58 (12/24 23:09)
※ 編輯: zptdaniel 來自: 118.233.32.58 (12/24 23:09)
→
12/24 23:11, , 9F
12/24 23:11, 9F
→
12/25 02:38, , 10F
12/25 02:38, 10F
→
12/25 08:10, , 11F
12/25 08:10, 11F
→
12/25 08:11, , 12F
12/25 08:11, 12F
→
12/25 08:11, , 13F
12/25 08:11, 13F
→
12/25 08:12, , 14F
12/25 08:12, 14F
→
12/25 08:13, , 15F
12/25 08:13, 15F
→
12/25 08:13, , 16F
12/25 08:13, 16F
討論串 (同標題文章)