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Zu (ii) Zun¨ achst ist (yn ) nach dem Lemma beschr¨ankt. Sei etwa |yn | < M f¨ ur alle n ∈ N. Sei > 0 gegeben. Dann gibt es ein N ∈ N, so dass f¨ ur alle n ≥ N gleichzeitig |xn − a| < und |yn − b| < . Dann gilt f¨ ur alle n ≥ N |xn yn − ab| = |(xn − a)yn + a(yn − b)| ≤ |xn − a| |yn | + |a| |yn − b| < (M + |a|). Zu (iii) Sei > 0 und dazu N ∈ N so gew¨ahlt, dass f¨ ur alle n ≥ N |xn − a| < , |yn − b| < und |yn − b| < Dann ist |yn | = |b + yn − b| ≥ |b| − |yn − b| ≥ |b| − 47 |b| . 2 |b| |b| = 2 2 und xn b − yn a |(xn − a)b + a(b − yn )| xn a = = − yn b yn b |yn b| |xn − a||b| + |a||yn − b| |b| + |a| ≤ .

22) Die letzte Ungleichung ist aber ¨ aquivalent zu 1 mit positivem α := 1 |x| < 1 |x| n = (1 + α)n f¨ ur alle n ≥ N . − 1. ) Nach der Bernoullischen Ungleichung (Beispiel 19) ist dann aber (1 + α)n = 1 + nα. W¨ahlt ur n ≥ N man also N > 1α , was nach dem Archimedischen Axiom ja m¨oglich ist, so folgt f¨ (1 + α)n ≥ 1 + nα ≥ 1 + N α > 1 + 1 > 1 und damit (22). Ende gut, alles gut! F¨ ur die Konvergenz einer Folge sind nach Definition nur die hinteren ” Glieder“ verantwortlich, was am Anfang passiert ist egal.

Aus lim inf xk = a = lim sup xk folgt also lim xk = a. h. a ist der einzige H¨aufngspunkt der Folge. Aus (i) folgt dann lim inf xk = a = lim sup xk . Lemma 97. Seien (xn ) eine nicht-negative Folge und a ∈ R. Dann gilt √ √ lim xn = a =⇒ lim xn = a. Beweis. Aus den Anordnungsaxiomen folgt f¨ ur nicht negative a, b a < b ⇐⇒ a2 < b2 , also 0 ≤ a < b =⇒ √ a< √ (30) b. ur alle n ≥ N . Dann ist 1. Fall: a√= 0. Sei > 0. Dann gibt es ein N ∈ N mit 0 ≤ xn < 2 f¨ √ √ | xn − 0| = xn < f¨ ur alle n ≥ N . Daraus folgt die Behauptung.

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