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) *一定要證明雙向 因為是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)類推（a） hint (c)可設計新的function h(x)=g(x)-f(x) 再套（a）的結 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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