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討論串[微積] 極限 & 連續
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Claim: f(x) = 0 for all x in [a,b]. ￣￣￣. Suppose not, there exists a number p in [a,b]. such that f(p) = m ≠ 0. Since f is continuous on [a,b], f is a
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Since f is continuous on [a,b]. then for all x in (R\Q)∩[a,b] implies. lim f(r) = f(x). r→x. make a sequence{r_n} which r_n→x as n→∞. each r_n is in Q
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看到這個敘述,我想到我看到得一題極限證明題:. Let f be continous on [a,b] and let f(x) = 0 when x is rational.. Prove that f(x) = 0 for every x in [a,b].. 嘗試去做證明:. Proof:.
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最近自己看交大的開放式課程(大一的微積分). 然後看到在講連續的地方有這個例題. f(x) = 0 , x is rational. 1 , x is irrational. 然後老師證明他不是連續的方法是這樣的. 假設f在x=2是連續 那麼. lim f(x) = 0 = f(2). x->2.
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