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I) lim f(x)=0 ==>lim I f(x) I= 0
x->c ................x->c

II) lim If(x) I=0 ==> lim f(x) = 0
x->c.......................x->c

2007-06-16 14:17:08 · 1 answers · asked by tc 1 in Science & Mathematics Mathematics

1 answers

Both of these statements are true.

Proof for 1: note that if [x→c]lim f(x) = 0, then [x→c]lim |f(x)| = |[x→c]lim f(x)| = |0| = 0, since the absolute value function is continuous.

For 2: Note that if [x→c]lim |f(x)| = 0, then ∀ε>0, ∃δ>0: 0<|x-c|<δ → ||f(x)|-0|<ε, but the absolute value function is idempotent, so ||f(x)|-0| = ||f(x)|| = |f(x)| = |f(x) - 0|, so this means ∀ε>0, ∃δ>0: 0<|x-c|<δ → |f(x)-0|<ε, which means [x→c]lim f(x) = 0.

2007-06-16 15:16:49 · answer #1 · answered by Pascal 7 · 0 0

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