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Daniel J Smith Custom TMUA alpha · Q17

Logic & proofFunctions & graphs

We say a real function f(x)f(x) is continuous at a point x0∈Rx_0\in\mathbb R if

for all ε>0\varepsilon>0 there exists δ>0\delta>0 such that for all x∈Rx\in\mathbb R, if ∣x−x0∣<δ|x-x_0|<\delta then ∣f(x)−f(x0)∣<ε|f(x)-f(x_0)|<\varepsilon

Which one of the following statements is true if and only if ff is not continuous at x0x_0?

AThere exists ε>0\varepsilon>0 such that for every δ>0\delta>0 there exists xx with ∣x−x0∣<δ|x-x_0|<\delta and ∣f(x)−f(x0)∣≥ε|f(x)-f(x_0)|\geq\varepsilon
BFor every ε>0\varepsilon>0 and every δ>0\delta>0 there exists xx with ∣x−x0∣<δ|x-x_0|<\delta and ∣f(x)−f(x0)∣≥ε|f(x)-f(x_0)|\geq\varepsilon
CThere exists ε>0\varepsilon>0 and δ>0\delta>0 such that for all xx with ∣x−x0∣<δ|x-x_0|<\delta we have ∣f(x)−f(x0)∣≥ε|f(x)-f(x_0)|\geq\varepsilon
DThere exists ε>0\varepsilon>0 such that for every δ>0\delta>0 we have ∣x−x0∣≥δ|x-x_0|\geq\delta or ∣f(x)−f(x0)∣≥ε|f(x)-f(x_0)|\geq\varepsilon
EFor every ε>0\varepsilon>0 and every δ>0\delta>0 there exists xx with ∣x−x0∣<δ|x-x_0|<\delta and ∣f(x)−f(x0)∣<ε|f(x)-f(x_0)|<\varepsilon
FThere exists δ>0\delta>0 such that for all ε>0\varepsilon>0 and all xx with ∣x−x0∣<δ|x-x_0|<\delta we have ∣f(x)−f(x0)∣<ε|f(x)-f(x_0)|<\varepsilon
GFor every ε>0\varepsilon>0 there exists δ>0\delta>0 such that there exists xx with ∣x−x0∣<δ|x-x_0|<\delta and ∣f(x)−f(x0)∣≥ε|f(x)-f(x_0)|\geq\varepsilon
HThere exists ε>0\varepsilon>0 such that for every δ>0\delta>0 if ∣x−x0∣<δ|x-x_0|<\delta then ∣f(x)−f(x0)∣≥ε|f(x)-f(x_0)|\geq\varepsilon

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