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James Chen TMUA Paper 2 · Q16

Paper 2Logic & proof

What is the negation of the following statement?

For all ϵ>0\epsilon>0, there exists δ>0\delta>0 such that if 0<∣x−a∣<δ0<|x-a|<\delta then ∣f(x)−L∣<ϵ|f(x)-L|<\epsilon.

AThere exists ϵ>0\epsilon>0, such that for all δ>0\delta>0, if 0<∣x−a∣<δ0<|x-a|<\delta then ∣f(x)−L∣≥ϵ|f(x)-L|\geq\epsilon.
BThere exists ϵ>0\epsilon>0, such that for all δ>0\delta>0, there exists 0<∣x−a∣<δ0<|x-a|<\delta such that ∣f(x)−L∣>ϵ|f(x)-L|>\epsilon.
CThere exists ϵ>0\epsilon>0, such that for all δ>0\delta>0, there exists 0<∣x−a∣<δ0<|x-a|<\delta such that ∣f(x)−L∣≥ϵ|f(x)-L|\geq\epsilon.
DThere exists ϵ>0\epsilon>0, such that for all δ>0\delta>0, if 0<∣x−a∣<δ0<|x-a|<\delta then ∣f(x)−L∣≥ϵ|f(x)-L|\geq\epsilon.
EThere exists ϵ>0\epsilon>0, such that for all δ≤0\delta\leq0, if 0<∣x−a∣<δ0<|x-a|<\delta then ∣f(x)−L∣<ϵ|f(x)-L|<\epsilon.
FThere exists ϵ>0\epsilon>0, such that for all δ≤0\delta\leq0, there exists 0<∣x−a∣<δ0<|x-a|<\delta such that ∣f(x)−L∣≥ϵ|f(x)-L|\geq\epsilon.

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