MaPhy-Ü8

MaPhy-Ü8 - Aufgabe 1 1 a) f (x) = x x (0, )...

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Aufgabe 1 a) f ( x ) = 1 x x (0 , ) Stetigkeit: ε > 0 δ ( ε,x 0 ) > 0 | | f ( x ) - f ( x 0 ) | < ε x U δ ( x 0 ) | {z } | x - x 0 | Sei ε > 0 ,x 0 > 0 beliebig, dann w¨ ahle δ := ε | x | 2 ε | x | +1 > 0 ⇒ ∀ x U δ ( x 0 ) : | f ( x ) - f ( x 0 ) | = ± ± ± 1 x - 1 x 0 ± ± ± = | x - x 0 | | x | | x 0 | < δ ( | x 0 | - δ ) | x 0 | | {z } ε = ε | x 0 | 2 ε | x 0 | +1 · 1 ± | x 0 |- ε | x 0 | 2 ε | x 0 | +1 ² | x 0 | = ε | x 0 | ( ε | x 0 | +1) | x 0 |- ε | x 0 | 2 = ε | x 0 | ε | x 0 | 2 + | x 0 |- ε | x 0 | 2 = ε gleichm¨aßige Stetigkeit: ε > 0 δ ( ε ) > 0 | | f ( x ) - f ( x 0 ) | < ε x,x 0 ∈ D | | x - x 0 | < δ Negation: ε > 0 δ > 0 x,x 0 ∈ D| | x - x 0 | < δ ∧ | f ( x ) - f ( x 0 ) | ≥ ε w¨ahle ε = 1, Sei δ > 0 beliebig, w¨ ahle x = δ,x 0 = δ 2 ⇒ | x - x 0 | = δ - δ 2 = δ 2 < δ aber | f ( x ) - f ( x 0 ) | = ± ± ± 1 δ - 1 δ/ 2 ± ± ± = 1 δ/ 2 > 1 δ < 1 f 0 stetig in x 0 f ( x + δ ) ---→ δ 0 f ( x ) ⇔ | f ( x + δ ) - f ( x ) | → 0 f gleichm¨aßig stetig in D ⇔ sup x ∈D
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MaPhy-&amp;Atilde;œ8 - Aufgabe 1 1 a) f (x) = x x (0, )...

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