推 chrisjon:喔喔喔!!看起來還真簡單orz 謝謝 01/17 00:55
※ 引述《dorminia (重新出發)》之銘言：
: ※ 引述《chrisjon (研究布丁狗)》之銘言：
: : Let f(x)=e^(-1/x^2) , x != 0 and f(0) = 0
: : Prove that f is twice differentiable at 0
: : 我的算法是f''(x)，然後再求lim x→0
: : 但是用羅必達法則微半天
: : 就是消不完~.~
: : 請各位先進幫忙一下，謝謝
f(x) - f(0) e^(-1/x^2) 1/x
f'(0) = lim ------------- = lim ------------- = lim ------------ (∞/∞型)
x→0 x - 0 x→0 x x→0 e^(1/x^2)
-1/x^2 x
= lim ------------------- = lim -------------- (0/∞) = 0 存在
x→0 (-2/x^3)*e^(1/x^2) x→0 2 e^(1/x^2)
f'(x) - f'(0) (2/x^3)*e^(-1/x^2)
f''(0) = lim --------------- = lim --------------------
x→0 x - 0 x→0 x
2/x^4
= lim --------------- (∞/∞型)
x→0 e^(1/x^2)
-8/x^5 4/x^2
= lim ------------------- = lim ------------ (∞/∞型)
x→0 (-2/x^3)*e^(1/x^2) x→0 e^(1/x^2)
-8/x^3 4
= lim ------------------- = lim ------------- = 0 存在
x→0 (-2/x^3)*e^(1/x^2) x→0 e^(1/x^2)
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※ 編輯: axis0801 來自: 61.62.121.187 (01/17 00:16)