Re: [微分] 很簡單
※ 引述《Honor1984 (希望願望成真)》之銘言:
※ 發信站: 批踢踢實業坊(ptt.cc)
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確實是這樣阿
不知道midarmyman昨天晚上後來有沒有去查導函數和導數相關的定義
f(x+h) - f(x)
導函數f'(x) = lim --------------
h->0 h
所以你給的函數f'(0) = 用上面這個方式求得的值 = 0
這是對的
其中f(0) = 0就是題目給的
但是有一點你要注意的是lim f(x) = 0 這個值0未必一定要是f(0)
x->0
這次題目剛好讓它=0是因為這樣就滿足lim f(x) = f(0)連續的性質
x->0
( 因為x^2 sin(1/x)在x=0的左右極限都剛好是0 )
f(x)就有連續的好性質
記得很多書上極限單元都會畫很多圖
告訴你什麼是極限值未必要等於函數在那點的值
這裡的意思就是這樣
另外2xsin(1/x) - cos(1/x)因為是在x^2 sin(1/x)有定義的區間微分
所以2xsin(1/x) - cos(1/x)在 x=0+ 與 x=0- 的極限代表lim f'(x) 與 limf'(x)
x->0+ x->0-
事實上前面一篇我也把左右極限的特性描述出來
是屆於在-1 ~ 1之間震盪得很厲害
不是1或者-1任何一個這樣單純的特定數值
但是同前面題到的的
lim f'(x) 或者 limf'(x) 未必要和f'(0)相等
x->0- x->0+
這代表f'(x)的導函數在x=0上不連續
然而f'(x) 在 x=0 處就只有一個值
那就是f'(0) (前面導函數代x=0的結果或者直接從導數定義做)
你不妨就把f'(x)看作是另外一個函數h(x)
再用上從前對於極限的定義
就不難理解
你甚至就把全部的函數都看作是一個點一個點的定義都沒關係
有時候這樣更能有效理解這題要問的重點
--
※ 發信站: 批踢踢實業坊(ptt.cc)
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