作者lockheart (Special Thanks to Eason)
看板Math
標題Re: [分析] 連續函數一題
時間Wed Aug 4 16:52:15 2010
※ 引述《sato186 (銀色轟炸機)》之銘言:
: ※ 引述《handsomecat3 (確定性的失落)》之銘言:
: : 設 f(x) is continuous on 開區間 (a,無限大) ,且
: : lim (f(x+1)-f(x)) = L
: : x->無限大
: : 則 lim f(x)/x = L (Cauchy)
: : x->無限大
: lim (f(x+1)-f(x)) = L, so for ε > 0,
: x→∞
: there exists an M such that
: L-ε < f(x+1)-f(x) < L+ε for x ≧ M. Then
: n(L-ε) + f(y) < f(y+n) < n(L+ε) + f(y)
: for fixed y ≧ M . Then
: f(y + n)
: lim ──── = L as ε→0.
: n→∞ n
We have for some fixed y ≧ M , L-ε < f(y+1)-f(y) < L+ε.
Similarly, L-ε < f(y+2)-f(y+1) < L+ε, etc.
Hence we can find that n(L-ε) < f(y+n)-f(y) < n(L+ε)
f(y + n)-f(y)
and this implies that lim ─────── = L.
n→∞ n
f(y + n)-f(y) f(y)
Since y is fixed and n→∞, L = L + 0 = lim ─────── + lim ───
n→∞ n n→∞ n
f(y + n)
= lim ────
n→∞ n
: Since f is continunous on (a,∞),
: f(x)
: lim ─── = L.
: x→∞ x
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◆ From: 180.177.32.37
※ 編輯: lockheart 來自: 180.177.32.37 (08/04 16:53)
→ lockheart :感覺有些地方跳太快,加上這些,有錯請多指教 :) 08/04 17:07
推 sato186 :正確無誤!! 08/04 17:10