作者steve1012 (steve)
看板NTUBIME-103
标题[闲聊] 2.4solution
时间Sat Sep 25 09:41:48 2010
2.4
35. A=4
37. B=\=3, A-B=3
49.
(1) if lim h->0 f(c+h)=f(c)
Givenε>0 exists δ>0 such that if |h|<δthen |f(c+h)-f(c)|< ε
Suppose know |x-c|<δ then by above |f(c+(x-c))-f(c)|< ε
=>|f(x)-fc |<, so lim x->c f(x)=f(c) by definition, which also means f
continues at x=c
(2)if f is continuous at c, lim x->c f(x)=f(c)
Givenε>0 exists δ>0 such that if |x-c|<δthen |f(x)-f(c)|< ε
Suppose |h|<δ => |(h+c)-c|<δ so |f(h+c)-f(c)|< ε
By definition lim h->0 f(h+c)=f(c)
*一定要证明双向 因为是iff*
50. (a) set 0<ε<f(c) by definition exists a δ>0 such that |x-c|<δ
(which also means x ∈ { c-δ,c+δ} )then |f(x)-f(c)|< ε
=> -ε<f(x)-f(c)< ε
=> 0 < f(c) -ε <f(x)
(b) (c)类推（a） hint (c)可设计新的function h(x)=g(x)-f(x) 再套（a）的结
果
56.
Set fe=[f(x)+f(-x)]/2
. fo =[f(x)-f(-x)]/2 then fe is an even function and fo is an odd
function
And fe +fo = f(x)
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1F：→ sky2857:大推呀~!!! 09/25 11:05
2F：推 stan999950:好棒好棒 09/25 22:20
3F：推 stan999950:你太扯写起来好顺利我好不容易弄好50. 09/27 00:23