연속의 정의
f f f a a a lim x → a f ( x ) = f ( a ) \lim_{x\to a} f(x) = f(a) lim x → a f ( x ) = f ( a ) a a a
f continuous at a ⟺ lim x → a f ( x ) = f ( a ) f \text{ continuous at } a \iff \lim_{x \to a} f(x) = f(a) f continuous at a ⟺ x → a lim f ( x ) = f ( a ) g ( a ) = a 2 + 1 ≠ a 2 = lim x → a g ( x ) g(a) = a^2 + 1 \neq a^2 = \lim_{x \to a} g(x) g ( a ) = a 2 + 1 = a 2 = x → a lim g ( x ) 예제
연속함수들의 합과 곱이 연속임을 증명하고, 이로부터 모든 다항식이 연속임을 보이세요.
∣ f ( x ) g ( x ) − f ( a ) g ( a ) ∣ ≤ ∣ f ( x ) ∣ ⋅ ∣ g ( x ) − g ( a ) ∣ + ∣ g ( a ) ∣ ⋅ ∣ f ( x ) − f ( a ) ∣ \lvert f(x)g(x) - f(a)g(a)\rvert \leq \lvert f(x)\rvert\cdot\lvert g(x) - g(a)\rvert + \lvert g(a)\rvert\cdot\lvert f(x) - f(a)\rvert ∣ f ( x ) g ( x ) − f ( a ) g ( a )∣ ≤ ∣ f ( x )∣ ⋅ ∣ g ( x ) − g ( a )∣ + ∣ g ( a )∣ ⋅ ∣ f ( x ) − f ( a )∣ a = 1 a = 1 a = 1 증명 보기
상수함수 c c c ∣ c − c ∣ = 0 < ε \lvert c - c\rvert = 0 < \varepsilon ∣ c − c ∣ = 0 < ε x ↦ x x \mapsto x x ↦ x ∣ x − a ∣ < δ \lvert x - a\rvert < \delta ∣ x − a ∣ < δ ∣ x − a ∣ < ε \lvert x - a\rvert < \varepsilon ∣ x − a ∣ < ε δ = ε \delta = \varepsilon δ = ε
(합) f , g f, g f , g a a a ε > 0 \varepsilon > 0 ε > 0 ∣ f ( x ) − f ( a ) ∣ < ε 2 \lvert f(x) - f(a)\rvert < \frac{\varepsilon}{2} ∣ f ( x ) − f ( a )∣ < 2 ε δ 1 \delta_1 δ 1 ∣ g ( x ) − g ( a ) ∣ < ε 2 \lvert g(x) - g(a)\rvert < \frac{\varepsilon}{2} ∣ g ( x ) − g ( a )∣ < 2 ε δ 2 \delta_2 δ 2 δ = min ( δ 1 , δ 2 ) \delta = \min(\delta_1, \delta_2) δ = min ( δ 1 , δ 2 ) ∣ ( f + g ) ( x ) − ( f + g ) ( a ) ∣ ≤ ∣ f ( x ) − f ( a ) ∣ + ∣ g ( x ) − g ( a ) ∣ < ε \lvert (f+g)(x) - (f+g)(a)\rvert \leq \lvert f(x) - f(a)\rvert + \lvert g(x) - g(a)\rvert < \varepsilon ∣( f + g ) ( x ) − ( f + g ) ( a )∣ ≤ ∣ f ( x ) − f ( a )∣ + ∣ g ( x ) − g ( a )∣ < ε
(곱) 항등식 ∣ f ( x ) g ( x ) − f ( a ) g ( a ) ∣ ≤ ∣ f ( x ) ∣ ⋅ ∣ g ( x ) − g ( a ) ∣ + ∣ g ( a ) ∣ ⋅ ∣ f ( x ) − f ( a ) ∣ \lvert f(x)g(x) - f(a)g(a)\rvert \leq \lvert f(x)\rvert\cdot\lvert g(x) - g(a)\rvert + \lvert g(a)\rvert\cdot\lvert f(x) - f(a)\rvert ∣ f ( x ) g ( x ) − f ( a ) g ( a )∣ ≤ ∣ f ( x )∣ ⋅ ∣ g ( x ) − g ( a )∣ + ∣ g ( a )∣ ⋅ ∣ f ( x ) − f ( a )∣ f f f a a a f f f ∣ x − a ∣ < δ 0 \lvert x - a\rvert < \delta_0 ∣ x − a ∣ < δ 0 ∣ f ( x ) − f ( a ) ∣ < 1 \lvert f(x) - f(a)\rvert < 1 ∣ f ( x ) − f ( a )∣ < 1 δ 0 \delta_0 δ 0 ∣ f ( x ) ∣ ≤ ∣ f ( a ) ∣ + 1 = : M \lvert f(x)\rvert \leq \lvert f(a)\rvert + 1 =: M ∣ f ( x )∣ ≤ ∣ f ( a )∣ + 1 =: M ∣ g ( x ) − g ( a ) ∣ < ε 2 M \lvert g(x) - g(a)\rvert < \frac{\varepsilon}{2M} ∣ g ( x ) − g ( a )∣ < 2 M ε δ 1 \delta_1 δ 1 ∣ f ( x ) − f ( a ) ∣ < ε 2 ( ∣ g ( a ) ∣ + 1 ) \lvert f(x) - f(a)\rvert < \frac{\varepsilon}{2(\lvert g(a)\rvert + 1)} ∣ f ( x ) − f ( a )∣ < 2 (∣ g ( a )∣ + 1 ) ε δ 2 \delta_2 δ 2 δ = min ( δ 0 , δ 1 , δ 2 ) \delta = \min(\delta_0, \delta_1, \delta_2) δ = min ( δ 0 , δ 1 , δ 2 ) ∣ f ( x ) g ( x ) − f ( a ) g ( a ) ∣ < M ⋅ ε 2 M + ∣ g ( a ) ∣ ⋅ ε 2 ( ∣ g ( a ) ∣ + 1 ) < ε 2 + ε 2 = ε \lvert f(x)g(x) - f(a)g(a)\rvert < M \cdot \frac{\varepsilon}{2M} + \lvert g(a)\rvert \cdot \frac{\varepsilon}{2(\lvert g(a)\rvert+1)} < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon ∣ f ( x ) g ( x ) − f ( a ) g ( a )∣ < M ⋅ 2 M ε + ∣ g ( a )∣ ⋅ 2 (∣ g ( a )∣ + 1 ) ε < 2 ε + 2 ε = ε
귀납법으로 x n x^n x n c x n c x^n c x n
∣ x ∣ \lvert x\rvert ∣ x ∣ 0 0 0 δ = ε \delta = \varepsilon δ = ε
∣ ∣ x ∣ − 0 ∣ = ∣ x ∣ < ε \left\lvert \lvert x\rvert - 0 \right\rvert = \lvert x\rvert < \varepsilon ∣ ∣ x ∣ − 0 ∣ = ∣ x ∣ < ε a = 0 a = 0 a = 0 증명 보기
보일 것은 ∣ x − 0 ∣ < δ ⇒ ∣ ∣ x ∣ − ∣ 0 ∣ ∣ < ε \lvert x - 0\rvert < \delta \Rightarrow \lvert \lvert x\rvert - \lvert 0\rvert \rvert < \varepsilon ∣ x − 0 ∣ < δ ⇒ ∣∣ x ∣ − ∣ 0 ∣∣ < ε ∣ ∣ x ∣ − 0 ∣ = ∣ x ∣ \lvert \lvert x\rvert - 0\rvert = \lvert x\rvert ∣∣ x ∣ − 0 ∣ = ∣ x ∣ ∣ x ∣ < ε \lvert x\rvert < \varepsilon ∣ x ∣ < ε
δ = ε \delta = \varepsilon δ = ε ∣ x − 0 ∣ = ∣ x ∣ < δ = ε \lvert x - 0\rvert = \lvert x\rvert < \delta = \varepsilon ∣ x − 0 ∣ = ∣ x ∣ < δ = ε ∣ x ∣ \lvert x\rvert ∣ x ∣ 0 0 0 [관점] 연속이 미분을 보장하지는 않습니다: ∣ x ∣ \lvert x\rvert ∣ x ∣ 0 0 0
f f f a a a g g g f ( a ) f(a) f ( a ) h = g ∘ f h = g \circ f h = g ∘ f a a a
∣ x − a ∣ < δ ⇒ ∣ f ( x ) − f ( a ) ∣ < η ⇒ ∣ g ( f ( x ) ) − g ( f ( a ) ) ∣ < ε \lvert x - a\rvert < \delta \Rightarrow \lvert f(x) - f(a)\rvert < \eta \Rightarrow \lvert g(f(x)) - g(f(a))\rvert < \varepsilon ∣ x − a ∣ < δ ⇒ ∣ f ( x ) − f ( a )∣ < η ⇒ ∣ g ( f ( x )) − g ( f ( a ))∣ < ε a = − 1 a = -1 a = − 1 ε ← η ← δ \varepsilon \leftarrow \eta \leftarrow \delta ε ← η ← δ 증명 보기
ε > 0 \varepsilon > 0 ε > 0 g g g b = f ( a ) b = f(a) b = f ( a ) ∣ y − b ∣ < η \lvert y - b\rvert < \eta ∣ y − b ∣ < η ∣ g ( y ) − g ( b ) ∣ < ε \lvert g(y) - g(b)\rvert < \varepsilon ∣ g ( y ) − g ( b )∣ < ε η > 0 \eta > 0 η > 0 이제 f f f η \eta η ∣ x − a ∣ < δ \lvert x - a\rvert < \delta ∣ x − a ∣ < δ ∣ f ( x ) − f ( a ) ∣ < η \lvert f(x) - f(a)\rvert < \eta ∣ f ( x ) − f ( a )∣ < η δ > 0 \delta > 0 δ > 0
∣ x − a ∣ < δ \lvert x - a\rvert < \delta ∣ x − a ∣ < δ ∣ f ( x ) − b ∣ < η \lvert f(x) - b\rvert < \eta ∣ f ( x ) − b ∣ < η ∣ g ( f ( x ) ) − g ( b ) ∣ < ε \lvert g(f(x)) - g(b)\rvert < \varepsilon ∣ g ( f ( x )) − g ( b )∣ < ε ∣ h ( x ) − h ( a ) ∣ < ε \lvert h(x) - h(a)\rvert < \varepsilon ∣ h ( x ) − h ( a )∣ < ε h h h a a a 수치 예시: g g g η = ε 2 \eta = \frac{\varepsilon}{2} η = 2 ε f f f η \eta η δ = η 2 \delta = \frac{\eta}{2} δ = 2 η δ = ε 4 \delta = \frac{\varepsilon}{4} δ = 4 ε
4. 연속함수